Free Fall Practice Problems for High Schools: Complete Guide
In this long article, we are going to practice some problems about a freely falling object in the absence of air resistance. All these questions are suitable for high school or college students, or even the AP Physics 1 exam.
Freely Falling Motion Problems
Problem (1): A tennis ball is thrown vertically upward with an initial speed of 17 m/s and caught at the same level above the ground.
(a) How high does the ball rise?
(b) How long was the ball in the air?
(c) How long does it take to reach its highest point?
Solution: Take up as the positive direction and the throwing point as the origin, so $y_0=0$.
(a) The ball goes up so high that its vertical velocity becomes zero. For this part of ascending motion, we can use the free fall kinematic equation $v^2-v_0^2=-2g(y-y_0)$. Substituting the known values into it and solving for $y$, we get \begin{gather*} v^2-v_0^2=-2g(y-y_0) \\\\ 0-17^2 = -2(10)(y_{max}-0) \\\\ \Rightarrow \boxed{y_{max}=14.45\,\rm m} \end{gather*}
(b) In all free fall practice problems, the best way to find the total flight time that the object was in the air is to use the kinematic equation $y-y_0=-\frac 12 gt^2+v_0t$. Then, substitute the coordinates of where the object landed on the ground into it.
As a rule of thumb, if the object returns to the same point of launch, its displacement vector is always zero, so $y-y_0=0$. Therefore, we have \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ 0=-\frac 12 (10)t^2+17t \\\\ \Rightarrow \boxed{5t^2-17t=0} \end{gather*} Solving this equation by factoring out the time and setting the remaining expression to zero, we get \begin{gather*} 5t^2-17t=0 \\\\ t(5t-17)=0 \\\\ \Rightarrow t=0 \quad ,\quad t=3.4\,\rm s \end{gather*} The first result corresponds to the initial time, and the other time, $\boxed{t_{tot}=3.4\,\rm s}$, is the amount of time the ball is in the air until it reaches the ground.
(c) At the highest point the vertical velocity is always zero, $v=0$. Using the equation $v=v_0-gt$, and solving for $t$, we have \begin{gather*} v=v_0-gt \\ 0=17-(10)t \\ \Rightarrow t_{top}=1.7\,\rm s \end{gather*} As you can see, the duration of the ball's going up, in the absence of air resistance, is always half the total flight time. Therefore, \[t_{top}=\frac 12 t_{tot}\]
Problem (2): A ball is dropped directly downward from a height of 45 meters with an initial speed of 6 m/s. How many seconds later does it strike the ground?
Solution: Taking up as the positive direction and the dropping point as the origin, we have $y_0=0$. Since the ball is moving downward, we choose a negative sign for its initial velocity, so $v_0=-6\,\rm m/s$.
The ball strikes the ground $45\,\rm m$ below the chosen origin, so its correct coordinate is $y=-45\,\rm m$.
The only kinematic equation that relates all these variables to the time is $y-y_0=-\frac 12 gt^2+v_0t$. Substituting the numerical values into this equation, yields \begin{gather*} y-y_0 =-\frac 12 gt^2+v_0t \\\\ -45-0 =-\frac 12 (10)t^2+(-6)t \\\\ \Rightarrow \boxed{5t^2+6t-45=0} \end{gather*} In the last step, after rearranging, we arrived at a quadratic equation, like $at^2+bt+c=0$, that its solution is found using the below formula \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \] where $a,b,c$ are constants. In this case, we have \[a=5, b=5, c=-45\] Substituting these values into the above formula, we get \begin{gather*} t=\frac{-5\pm\sqrt{5^2-4(5)(-45)}}{2(5)} \\\\ t=2.46\,\rm s \quad , \quad t=-3.66 \end{gather*} Since we choose the positive time for free fall problems, the ball would reach the ground approximately $2.5\,\rm s$ after being dropped.
Problem (3): A coin is tossed vertically upward and remains $\rm 0.6\, s$ in the air before it comes back down.
(a) How fast is the coin tossed?
(b) How high does the coin rise?
Solution: Let the tossing point be the origin, so $y_0=0$. The time duration the coin is in the air until it reaches the highest point is $t=0.6\,\rm s$. Recall that in all free fall problems, the object at the highest point has a velocity $v=0$.
(a) Use the kinematic equation $v=v_0-gt$, substitute the above-known values, and solve for the unknown initial speed $v_0$ \begin{align*} v&=v_0-gt \\ 0&=v_0-(10)(0.6) \\ \Rightarrow v_0&=6\,\rm m/s \end{align*} Therefore, the coin was tossed vertically upward with an initial speed of $6\,\rm m/s$.
(b) In this part, we need a kinematic equation that relates the distance traveled to the time taken, $y-y_0=-\frac 12 gt^2+v_0t$, or to the initial and final velocities, $v^2-v_0^2=-2g(y-y_0)$. We want to use the second equation as follows. \begin{align*} v^2-v_0^2&=-2g(y-y_0) \\\\ (0)^2-(60)^2 &= -2(10)(y-0) \\\\ y&=\frac{-6^2}{-20} \\\\ &=\boxed{1.8\,\rm m} \end{align*} You can also use the first equation and arrive at the same result. Check it out yourself.
Problem (4): A small stone is shot straight up with an initial speed of $\rm 15\,m/s$.
(a) How long does the stone take to reach its maximum height?
(b) How high does the stone go?
(c) With what speed would the stone hit the ground?
Solution: The known data is $y_0=0$ and $v_0=15\,\rm m/s$.
(a) When an object is thrown vertically upward in the air, it rises until it reaches a point where its vertical velocity becomes zero; otherwise, it continues on its way. So, in this part, at the maximum height, we have $v=0$. Substituting all these known numerical values into the equation $v=v_0-gt$, and solving for the unknown time $t$, we get \begin{align*} 0 &= 15-(10)t \\ \Rightarrow t&=\boxed{1.5\,\rm s} \end{align*} Thus, it takes $1.5\,\rm s$ for the stone to reach the highest point.
(b) For this part, we have the time taken to reach the maximum height $t=1.5\,\rm s$, the initial and final velocities. Thus, we can use either the equation $y-y_0=-\frac 12 gt^2+v_0t$ or $v^2-v_0^2=-2g(y-y_0)$. We chose the second equation. \begin{align*} 0-(15)^2 &= -2(10)(y_{max}-0) \\\\ y_{max} &=\frac{-15^2}{-20} \\\\ \Rightarrow y_{max}&= \boxed{11.25\,\rm m} \end{align*} Thus, the stone reaches the maximum height of $11.25\,\rm m$.
(c) When the stone hits the ground at the same level as the throwing point, then according to the definition of displacement, it displaces nothing. This means that the displacement between initial and final points $\Delta y=y-y_0$ is zero.
Using this fact, we can use the equation $v^2-v_0^2=-2g(y-y_0)$ to get the velocity at the moment of hitting the ground. \begin{align*} v^2-(15)^2 &=-2(10)(0) \\ \Rightarrow v^2 = 15^2 \end{align*} Taking the square root of both sides, yields two roots of $v=\pm 15\,\rm m/s$. Recall that velocity is a vector in physics that has both magnitude and direction.
The stone is hitting the ground, so the correct sign is a negative for its velocity, i.e., $v=-15\,\rm m/s$.
Problem (5): From the top of a high cliff, a stone is released. It is seen after $2.5\,\rm s$, the stone strikes the ground. How high is the cliff?
Solution: The stone is released, so its initial speed is zero, $v_0=0$. The total flight time is also $t=2.5\,\rm s$. Taking the releasing point as the origin, we will have $y_0=0$. In the kinematic equations, there is vertical displacement in only two of them, i.e., $v^2-v_0^2=-2g(y-y_0)$ and $y-y_0=-\frac 12 gt^2+v_0t$. As you can see, to use the first equation, we must have the final velocity where the stone hits the ground, and for the second equation, the total flight time is needed. Hence, it is simpler to apply the second equation and solve for the unknown vertical distance $y$. \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ y-0=-\frac 12 (10)(2.5)^2+(0)(2.5) \\\\ \Rightarrow \quad \boxed{y=-31.25\,\rm m} \end{gather*} The negative indicates that the stone strikes the ground $31.25\,\rm m$ below our chosen origin.
Problem (6): A stone is released at rest from a height and falls freely for $4\,\rm s$.
(a) What is the stone's velocity $1.5\,\rm s$ after releasing?
(b) How high is the height?
Solution: The stone is released at rest, so $v_0=0$. Take upward as the positive direction. The total flight time is $t_{tot}=4\,\rm s$.
(a) With this known information, use the equation $v=v_0-gt$ to find the stone's velocity at each instant of time. \begin{gather*} v=v_0-gt \\ v=0-(10)(1.5) \\ \Rightarrow \boxed{v=-15\,\rm m/s} \end{gather*} The negative indicates the direction of the stone at that moment, which is facing down.
(b) The time between releasing the stone and hitting the ground is given. With this known information, use the equation $y-y_0=-\frac 12 gt^2 +v_0t$ and solve for the total distance fallen by the stone. \begin{gather*} y-0=-\frac 12 (10)(4)^2+(0)(4) \\\\ \Rightarrow \quad \boxed {y=-80\,\rm m} \end{gather*} The minus sign reminds us that the stone strikes the ground $80\,\rm m$ below our chosen origin.
Problem (7): A person throws a light stone straight up and catches it $2.6\,\rm s$ later. With what speed did he throw the stone, and to what height does the stone go up?
Solution: As with any other free-fall problem, let upward be the positive direction and the throwing point be the origin of the coordinate system so that $y_0=0$.
The stone is caught at the same level of throwing, so its vertical displacement is zero, $\Delta y=y-y_0=0$. The total flight time is also known. Hence, applying the vertical displacement kinematic equation, $y-y_0=-\frac 12 gt^2+v_0t$, and solving for the initial speed $v_0$ gives us \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ 0=-\frac 12 (10)(2.6)^2+v_0 (2.6) \end{gather*} Factoring out $2.6$ from the last expression, would get \begin{gather*} (2.6)(-5(2.6)+v_0)=0 \\\\ \Rightarrow \quad \boxed{v_0=13\,\rm m/s} \end{gather*} As a side note, if an object is thrown vertically upward and caught at the same level, by knowing the time interval between these two moments, we can use the following formula to find its initial speed \[v_0=\frac 12 g t\] The stone goes up until it reaches a point where its vertical velocity is zero, i.e., $v=0$. Now that we know the stone's initial speed, applying the equation $v^2-v_0^2=-2g(y-y_0)$ gives us \begin{gather*} 0-(13)^2=-2(10)(y-0) \\\\ \Rightarrow \quad \boxed{y=8.45\,\rm m} \end{gather*}
Problem (8): A baseball is thrown straight up with a speed of $25\,\rm m/s$.
(a) With what speed is it moving when it is at a height of $10\,\rm m$?
(b) How much time does it take to reach that point?
Solution: As usual, take up the positive $y$-direction and set $y_0=0$. The initial speed is also $v_0=25\,\rm m/s$.
(a) The only time-independent kinematic equation that relates these known values to each other is $v^2-v_0^2=-2g(y-y_0)$. Substituting the known numerical values into this, we get \begin{gather*} v^2-(25)^2=-2(9.8)(10-0) \\\\ v^2=429 \\\\ \Rightarrow \quad v=\pm 20.7\,\rm m/s \end{gather*} As you can see, we obtained a speed with two different signs. Here, the speed with a positive sign indicates that the baseball at that desired height is being moved upward, while the negative sign hints to us that the baseball is at that height when it is moving down.
Thus, we arrive at the fact that the baseball reaches that height twice, once when it is going up and once when it is going down.
(b) In the previous part, we found out that we were at that height twice. Thus, there are two corresponding times for this situation. Applying the equation $y-y_0=-\frac 12 gt^2+v_0t$ and solving for the required time $t$, we get \begin{gather*} 10-0=-\frac 12 (10)t^2+25t \\\\ \Rightarrow \quad 4.9t^2-25t+10=0 \end{gather*} The quadratic equation above has two solutions as below \begin{gather*} t_1=0.44\,\rm s \quad , \quad t_2=4.66\,\rm s \end{gather*} the first time, $t_1$ is when the ball is going up, and the second one corresponds to when it is moving down.
Note: The quadratic equation of $at^2+bt+c=0$ has the following solution formula \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
Problem (9): A helicopter is ascending vertically upward with a constant speed of $6.5\,\rm m/s$ and carrying a package of $2\,\rm kg$. When it reaches a height of $150\,\rm m$ above the surface, it drops the package. How much time does it take for the package to hit the surface?
Solution: Take up the positive direction, as always. At the moment that the package is released, it has the speed of its carrier.
There is a subtle point in this case. Since the object is moving down, in the opposite direction of our positive direction, a negative must accompany it. Therefore, the correct input for the initial speed is $v_0=-6.5\,\rm m/s$.
Applying the equation $y-y_0=-\frac 12 gt^2+v_0t$, we will have \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ -150-0=-\frac 12 (10)t^2+(-6.5)t \\\\ \Rightarrow 5t^2+6.5t-150=0 \end{gather*} Using a graphing calculator or the formula in the previous question, we find that it takes about $4.9\,\rm s$ for the package to reach the ground.
Note that in the above, we inserted a negative for the vertical height as $y=-150\,\rm m$, since the package hits the ground $150\,\rm m$ below our chosen origin.
Problem (10): A stone is thrown vertically upward from a building $15\,{\rm m}$ high with an initial velocity of $10\,{\rm m/s}$. What is the stone's velocity just before hitting the ground?
Solution: In all freely falling practice problems, the important note is choosing the origin. Usually, the throwing (releasing or dropping) point is the best choice. In this case, the hitting point is below the origin, so its vertical displacement ($y$) is negative.
Applying the time-independent free fall kinematic equation, we have \begin{align*}v_f^{2}-v_i^{2}&=2\,(-g)\Delta y\\v_f^{2}-(10)^{2}&=2\,(-10)(-15) \\ \Rightarrow v_f&=\pm 20\,{\rm m/s}\end{align*} Since velocity is a vector quantity and just before striking the ground its direction is vertically downward, the negative value must be chosen, i.e., $v_f=-20\,{\rm m/s}$.
Problem (11): A bullet is fired vertically downward with an initial speed of $15\,{\rm m/s}$ from the top of a tower of $20\,{\rm m}$ high. What is its velocity at the instant of striking the ground?
Solution: Let the origin be the firing point. Using the below kinematic equation, we have \begin{align*}v_f^{2}-v_i^{2}&=-2g(y-y_0) \\\\ v_f^{2}-0&=-2(10)(-20)\\\Rightarrow v_f&=\pm 20\,{\rm m/s}\end{align*} Since the direction of velocity at the moment of striking the ground is downward, we must choose the negative correct sign, so we have $v_f=-20\,\rm m/s$.
Problem (12): There is a well with a depth of $34\,{\rm m}$. A person drops a stone vertically into it with an initial velocity of $7\,{\rm m/s}$. What is the time interval between dropping the stone and hearing its impact sound? (Assume $g=10\,{\rm m/s^2}$ and the speed of the sound in the air is $340\,{\rm m/s}$).
Solution: This motion has two parts. One is descending into the well, which is a constant acceleration motion, and the other is ascending to the sound of impact, which is a uniform motion with constant speed.
For the first part, use the kinematic equation $y-y_0=-\frac 12 gt^{2}+v_0 t$ to find the falling time as \begin{gather*}y-y_0=-\frac 12 gt^{2}+v_0 t \\\\ -34-0=-\frac 12 (10)t^{2}+(-7)t \\\\ \Rightarrow 5t^{2}+7t-34=0 \end{gather*} Since initial velocity is a vector and its direction is initially downward, a negative is included in front of it. The above quadratic equation has two solutions: $t_1=2\,{\rm s}$ and $t_2=-3.4\,{\rm s}$. Obviously, a negative value for time is not accepted because time in all kinematic questions must be a positive quantity.
The second part is a uniform motion as the speed of sound is constant, so we can calculate its rising time using the definition of average velocity: \begin{align*} t&=\frac{\Delta y}{v} \\\\ &=\frac{34}{340}=0.1\,{\rm s}\end{align*} Thus, the total time is obtained as $t_T=2+0.1=2.1\,{\rm s}$.
Problem (13): A stone is dropped vertically downward from the top of a building with a height of $h$. What is its speed at the height of $\frac h2$?
Solution: Let the dropping point be the origin. Thus, in the kinematic equations, the vertical displacement must be negative, i.e., $y-y_0=-\frac h2$. Use the below equation to find the speed at the desired level \begin{align*}v_2^2-v_1^2&=-2g(y-y_0) \\\\ v_2^2 -0&=-2g(-\frac h2) \\\\ \Rightarrow v_2&=\sqrt{gh}\end{align*}
Problem (14): You throw a baseball straight up in the air and catch it 3.4 seconds later at the same place at which you threw it. How high did it go? (DIFFICULT)
Solution: First of all, choose a coordinate system with the upward direction as positive. In the absence of air resistance, when you throw an object upward, the time it takes to rise is equal to the time it takes to fall. In other words, the time of reaching the highest point is half the total time the object is in the air \[t_{rise}=t_{fall}=\frac 12 t_{tot}\] Other than the total time of flight, no other relevant information is given. Substitute the time taken to reach the highest point into $v=v_0-gt$ and solve for the initial velocity $v_0$ as below \begin{gather*} v_{top}=v_0-gt_{rise} \\\\ 0=v_0-(9.8)(1.7) \\\\ \Rightarrow v_0=16.6\,\rm m/s\end{gather*} Note that at the highest point $v_{top}=0$.
Now, apply the equation $v^2-v_0^2=-2g\Delta y$ between the initial point and the highest point so that $\Delta y=h$ and solve for $h$. \begin{gather*} v^2_{top}-v_0^2=-2gh \\\\ (0)^2-(16.6)^2=-2(9.8)h \\\\ \Rightarrow \boxed{h=14.0\,\rm m} \end{gather*} Therefore, the ball rises as high as $14.0\,\rm m$.
Problem (15): A bullet is fired vertically upward from a height of $90\,{\rm m}$ and after $10\,{\rm s}$ reaches the ground. After $2\,{\rm s}$ from the throwing point, the bullet is how far away from the surface? ($g=9.8\,{\rm m/s^2}$)
Solution: In a free-fall problem, the vertical position of an object in an instant of time is given by the kinematic equation $y=-\frac 12 gt^{2}+v_0 t+y_0$. Let the firing point be the origin, so $y_0=0$. To complete the equation above, you must have an initial speed.
To find the initial speed, apply the kinematic formula $y=-\frac 12 gt^{2}+v_0 t+y_0$ between the origin and striking point. \begin{gather*}y-y_0=-\frac 12 gt^{2}+v_0 t \\\\ -90-0=-\frac 12 (9.8)(10)^{2}+v_0(10) \\\\ \Rightarrow \boxed{v_0=40\,\rm m/s}\end{gather*} Notice that the striking point is $-90\,{\rm m}$ below the origin that's why the negative is entered for the vertical displacement $y$.
Having the initial velocity, substitute it into the above equation again but with time $t=2\,{\rm s}$ to find the position of the bullet after $2\,{\rm s}$ with respect to the firing point \begin{align*}y-y_0&=-\frac 12 gt^{2}+v_0 t\\\\ &=-\frac 12\,(9.8)(2)^{2}+(40)(2) \\\\ \Rightarrow y&=\boxed{60.4\,\rm m} \end{align*}
Problem (16): A ball is thrown vertically upward with an initial velocity of $18\,\rm m/s$. How many seconds after throwing, the ball's speed is $9\,\rm m/s$ downward?
Solution: In all kinematic equations, $x,y,v,v_0$, and $a$ are vectors, so their signs matter. A speed of $9\,{\rm m/s}$ downward means a velocity of $-9\,{\rm m/s}$. Downward or upward indicates the direction of velocity.
Take up as the positive $y$ direction. Now use the equation $v=v_0-gt$ to find the velocity at any later time.\begin{gather*}v=v_0-gt\\-9=+18-(10)t\\ \Rightarrow \quad \boxed{t=2.7\,\rm s}\end{gather*}
Problem (17): An object is thrown vertically upward in the air from a $100\,{\rm m}$ height with an initial velocity of $v_0$. After $5\,{\rm s}$, it reaches the ground. Determine the magnitude and direction of the initial velocity.
Solution: Let the origin be the throwing point (so $y_0=0$) and the upward direction positive. Substitute the given total time into the vertical displacement kinematic equation $y-y_0=-\frac 12 gt^2+v_0t$ with $y-y_0=-100\,{\rm m}$ (since the impact point is below the origin). \begin{gather*} y-y_0 =-\frac 12 gt^2+v_0t \\\\ -100=-\frac 12\,(10)(5)^2+v_0 (5) \\\\ \Rightarrow \boxed{v_0=-5\,\rm m/s}\end{gather*} the minus sign indicates the initial velocity is downward with the magnitude (speed) of $5\,\rm m/s$.
Problem (18): From a height of $15\,{\rm m}$, a ball is kicked vertically up into the air with an initial speed of $v_0$. It reaches the highest point of its path with an elevation of $20\,{\rm m}$ from the surface. Find the initial velocity $v_0$.
Solution: The highest point is $5\,{\rm m}$ above the kicking point. Apply the time-independent kinematic equation below to find the initial velocity \begin{gather*}v^2-v_0^2=-2g(y-y_0) \\\\ 0-v_0^2=-2(10)(5) \\\\ \Rightarrow v_0=\pm 10\,{\rm m/s}\end{gather*} Because the ball kicked upward so we must choose the plus sign, i.e. $v_0=+10\,{\rm m/s}$. In the above, we used the fact that in all freely falling problems, at the highest point (apex) the velocity is zero ($v=0$).
Recall that projectiles are a particular type of freely falling motion with a launch angle of $\theta=90$ with its own formulas.
Problem (19): From the bottom of a $25\,{\rm m}$-depth well, a stone is thrown vertically upward with an initial speed of $30\,{\rm m/s}$.
(a) How high does the stone rise out of the well?
(b) Before the stone returns into the well, how many seconds are outside the well?
Solution:
(a) Let the bottom of the well be the origin, so $y_0=0$. First, we find how much distance the ball rises. Recall that the highest point is where $v=0$, so we have \begin{gather*}v^2-v_0^2=-2g(y-y_0) \\\\ 0-(30)^2=-2(10)(y-0) \\\\ \Rightarrow \boxed{y=45\,\rm m}\end{gather*} Of this height, $25\,{\rm m}$ is for the well's depth, so the stone is $20\,{\rm m}$ outside of the well.
(b) We want to examine the duration between exiting and reentering the stone into the well. During this time interval, the ball returns to its initial position, so its displacement vector is zero, i.e., $\Delta y=y-y_0=0$. If we want to use the equation, $y-y_0=-\frac 12 gt^2+v_0t$, the speed of the stone is required exactly when it leaves the well.
The speed at the bottom of the well is known. Thus, apply the equation $v^2-v_0^2=-2g\Delta y$ to find the speed just before it leaves the well. \begin{gather*}v^2-v_0^2=-2g\Delta y \\\\ v^2-(30)^{2}=-2(10)(25) \\\\ \Rightarrow v=+20\,{\rm m/s}\end{gather*} This speed can be used as the initial speed for the part where the stone is outside the well. Hence, the total time the stone is out of the well is obtained as below \begin{gather*} \Delta y=-\frac 12 gt^{2}+v_0 t\\ 0=-\frac 12 (10)t^{2}+(20)(2) \end{gather*} Solving for $t$, one can obtain the required time as $t=4\,{\rm s}$.
Problem (20): From the top of a $20-{\rm m}$-high tower, a small ball is thrown vertically upward. If $4\,{\rm s}$ after throwing, it hit the ground. How many seconds before striking the surface does the ball again meet the original throwing point? (Air resistance is neglected and $g=10\,{\rm m/s^2}$).
Solution: Let the origin be the throwing point. The ball strikes the ground $20\,\rm m$ below our chosen origin, so its total displacement between initial position $y_i=0$ and final position is $\Delta y=y_f-y_i=-20\,\rm m$. The total time in which the ball is in the air is also $4\,{\rm s}$. With these known values, one can find the initial velocity as \begin{gather*}\Delta y=-\frac 12 gt^{2}+v_0t \\\\ -25=-\frac 12 (10)(4)^{2}+v_0(4) \\\\ \Rightarrow \boxed{v_0=15\,\rm m/s} \end{gather*} When the ball returns to its initial position, its total displacement is zero, i.e., $\Delta y=0$ so we can use the following kinematic equation to find the total time it takes the ball to return to the starting point \begin{gather*}\Delta y=-\frac 12 gt^{2}+v_0t \\\\ 0=-\frac 12 (10)t^{2}+(15)t \end{gather*} Rearranging and solving for $t$, we get $t=3\,{\rm s}$.
Problem (21): A rock is thrown vertically upward in the air. It reaches the height of $40\,{\rm m}$ from the surface at times $t_1=2\,{\rm s}$ and $t_2$. Find $t_2$ and determine the greatest height reached by the rock (neglect air resistance and assume $g=10\,{\rm m/s^2}$).
Solution: Let the throwing point (surface of the ground) be the origin. Between our chosen origin and the point with known values $h=4\,{\rm m}$, $t=2\,{\rm s}$ one can write down the kinematic equation $\Delta y=-\frac 12 gt^{2}+v_0\,t$ to find the initial velocity as \begin{gather*} \Delta y=-\frac 12 gt^{2}+v_0t \\\\ 40=-\frac 12 (10)(2)^2 +v_0(2) \\\\ \Rightarrow v_0=30\,{\rm m/s}\end{gather*} Now we are going to find the times when the rock reaches the height $40\,{\rm m}$ (Recall that when an object is thrown upward, it passes through every point twice). Applying the same equation above, we get \begin{gather*} \Delta y=-\frac 12 gt^{2}+v_0t \\\\ 40=-\frac 12\,(10)t^2+30t \\\\ \Rightarrow \boxed{5t^2-30t+40=0} \end{gather*} Rearranging and solving for $t$ using quadratic equation formula, two times are obtained i.e. $t_1=2\,{\rm s}$ and $t_2=4\,{\rm s}$. Thus, again after $4\,\rm s$ the ball again is at the same height of $40\,\rm m$ from the surface.
The greatest height is where the vertical velocity becomes zero so we have \begin{gather*}v^2-v_0^2 =-2g\Delta y \\\\ 0-(30)^2=-2(10)\Delta y\\\\ \Rightarrow \boxed{\Delta y=45\,\rm m}\end{gather*} Thus, the highest point where the rock can reach is located at $H=45\,{\rm m}$ above the ground.
Problem (22): A ball is launched with an initial velocity of $30\,{\rm m/s}$ straight upward. How long will it take the ball to reach $20\,{\rm m}$ below the highest point for the first time? (neglect air resistance and assume $g=10\,{\rm m/s^2}$).
Solution: Between the origin (surface level) and the highest point ($v=0$) apply the time-independent kinematic equation below to find the greatest height $H$ where the ball reaches.\begin{gather*}v^2-v_0^2=-2g\Delta y \\\\ 0-(30)^2=-2(10)H \\\\ \Rightarrow \boxed{H=45\,\rm m}\end{gather*} This is the maximum height that the ball can reach. The $20\,{\rm m}$ below this maximum height $H$ has a height of $h=45-20=25\,{\rm m}$. Now use the vertical displacement kinematic equation between the throwing point and the desired position to find the required time taken. \begin{gather*} \Delta y=-\frac 12 gt^2+v_0t \\\\ 25=-\frac 12(10)t^2+30(t)\end{gather*} Solving for $t$ (using quadratic formula), we get $t_1=1\,{\rm s}$ and $t_2=5\,{\rm s}$ one for up way and the second for down way.
Problem (23): A stone is launched directly upward from the surface level with an initial velocity of $20\,{\rm m/s}$. How many seconds after launch is the stone's velocity $5\,{\rm m/s}$ downward?
Solution: Let the origin be at the surface level and take the positive direction up. Therefore, we have initial velocity $v_0=+20\,{\rm m/s}$ and final velocity $v=-5\,{\rm m/s}$. Use the velocity kinematic equation $v=v_0-gt$ to find the desired time as below \begin{gather*}v=v_0-gt \\-5=+20-10\times t \\ \Rightarrow \boxed{t=2.5\,\rm s}\end{gather*}
Problem (24): From a $25-{\rm m}$ building, a ball is thrown vertically upward at an initial velocity of $20\,{\rm m/s}$. How long will it take the ball to hit the ground?
Solution: Origin is considered to be at the throwing point, so $y_0=0$. Apply the position kinematic equation below to find the desired time \begin{gather*} y-y_0=-\frac 12 gt^2+v_0 t \\\\ -25=-\frac 12(10)t^2+20t \\\\ \Rightarrow 5t^2-20t-25=0 \end{gather*} Rearranging and converting it into the standard form of a quadratic equation $at^2+bt+c=0$, its solutions are obtained as \begin{align*}t_{1,2}&=\frac{-b\pm \sqrt{b^2-4\,ac}}{2a} \\\\ &=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-5)}}{2(1)}\\\\ &=-1 \, \text{and} \, 5 \end{align*} Therefore, the time needed for the ball to hit the ground is $5\,{\rm s}$.
Problem (25): From the top of a building with a height of $60\,{\rm m}$, a rock is thrown directly upward at an initial velocity of $20\,{\rm m/s}$. What is the rock's velocity at the instant of hitting the ground?
Solution: Apply the time-independent kinematic equation as \begin{gather*} v^2-v_0^2 =-2g(y-y_0) \\\\ v^2-(20)^2 =-2(10)(-60) \\\\ v^2 =1600\\\\ \Rightarrow \quad \boxed{v=\pm 40\,\rm m/s} \end{gather*} Therefore, the rock's velocity when it hit the ground is $v=-40\,{\rm m/s}$.
In this article, you learned how to solve free fall problems step-by-step using simple kinematic equations.
Author: Dr. Ali Nemati
Published: 8/11/2022
