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# Formula for Projectile Motion with Examples for High Schools

## Definition of projectile motion:

Any object that is thrown into the air with an angle $\theta$ is projectile and its motion called projectile motion. In other words, any motion in two dimensions and only under the effect of gravitational force is called projectile motion.

## Formula for Projectile Motion:

The following are the complete projectile motion equations in vertical and horizontal directions.

### In horizontal direction:

\begin{align*} \text{Displacement}&:\,\Delta x=\underbrace{\left(v_0 \cos \theta\right)}_{v_{0x}}t\\ \text{Velocity}&:\, v_x=v_0 \cos \theta \end{align*}

### In vertical direction:

\begin{align*} \text{Displacement}&:\, \Delta y=\frac 12 (-g)t^2 +(\underbrace{v_0 \sin \theta}_{v_{0y}})\,t\\ \text{Velocity I}&:\, v_y = \underbrace{v_0 \sin \theta}_{v_{0y}}+(-g)t \\ \text{Velocity II}&:\, v_y^2 -\left(v_0 \sin \theta\right)^2=2(-g)\Delta y \end{align*}

#### Formula for object's speed in projectile motion (Using the Pythagorean theorem):

$v=\sqrt{v_x^2 + v_y^2 }$

#### Formula for angle of velocity vector at any time with x-axis:

\begin{align*} \theta &= \tan^{-1} \left(\frac{v_y}{v_x}\right)\\ &=\tan^{-1} \left(\frac{v_0 \sin \theta -gt}{v_0 \cos \theta}\right) \end{align*}

#### Formula for path of projectile (Trajectory path equation):

$y(x)=x\:\tan \theta-\frac{gx^2}{2v_0^2\,\cos^2 \theta}$

Where, $y$ and $x$ are the vertical and horizontal displacement, respectively. $\theta$ is the angle with the horizontal and $v_0$ is the initial speed.

#### Horizontal Range of a Projectile:

The horizontal distance between launch and striking points is called the Range of Projectile whose equation is
$R= \frac{v_0^2}{g}\,\sin 2\theta$

#### Total Time of Flight for a Projectile:

The total time of a projectile in the air is calculated as
$t=\frac{2v_0 \sin \theta}{g}$

#### Formula for maximum height reached by a projectile:

$H=\frac{v_0^2 \sin^2 \theta}{2g}$

#### Horizontal Projectile Motion Formula:

All the above formulas were based on the non-zero launch angle. If the projectile is thrown in the air at an angle of $\theta=0$, then there is no $y$-component of the initial velocity i.e. $v_{0y}=0$.
This case is called horizontal projectile motion and its formulas are as below
\begin{align*} v_{0x}&=v_0 \\ v_{0y}&=0\\ \Delta x&=v_0 \,t\\ \Delta y&=-\frac 12 g\,t^2\\
v_x&=v_{0x}\\ v_y&=-gt\\ v^{2} &= -2\,g\,\Delta y \end{align*} See Example (3) below.

Use formulas for projectile motions to practice the following examples.

### Projectile Motion Solved Examples:

Example (1):

A projectile is fired at $150\,{\rm m/s}$ from a cliff with a height of $200\,{\rm m}$ at an angle of $37^\circ$ from horizontal. Find the following:
(a) the distance at which the projectile hit the ground.
(b) the maximum height above the ground reached by the projectile.
(c) the magnitude and direction of the projectile velocity vector at the instant of impact to the ground.

Solution:

Let the firing point be the origin of coordinate, $y$ is positive upward and $x$ is positive to the right. Since the projectile hit the ground below the considered origin so its coordinate is $(x=?,y=-200\,{\rm m})$.

(a) The formula for horizontal distance of a projectile is given by $\Delta x=(v_0\,\cos \theta)\, t$, since we are asked to find the total distance from launching to striking point $(x=?,y=-200\,{\rm m})$, which is the range of projectile, so the total time of flight is required which is obtained as below \begin{align*} y&=-\frac 12 g\,t^2 + (v_0 \sin \theta)\,t +y_0 \\ -200&=-\frac 12 (9.8)\,t^2 + (150\times\,\sin 37^\circ)\,t+0\\ &\Longrightarrow \ 4.8\,t^2-90\,t-200=0 \end{align*} The solution to the quadratic equation $ax^2 + bx+c=0$ is given $x=\frac{-b\pm\sqrt{b^2 - 4\,a\,c}}{2a}$ Using that, one can find the total time of flight as below \begin{align*}
t&=\frac{90\pm\sqrt{(90)^2-4\,(4.8)(-200)}}{2(4.8)}\\ &= t_1=20.757\,{\rm s} \ \ \text{and} \ \ t_2\approx-2\,{\rm s} \end{align*} Because the projectile fired at $t=0$ so the time of strike can not be a negative value. Thus, the total time is $t=20.757\,{\rm s}$. Now substitute it into the horizontal distance formula to find the RANGE of Projectile as below
\begin{align*} \Delta x&=(v_0\,\cos \theta)\, t \\ &=(150\times \,\cos 37^\circ)(20.757)\\
&=(150\times 0.8)(20.757)\\ &=95.83\,{\rm m} \end{align*}

(b) One of the key features of projectile motions is that its vertical velocity, $v_y$, at the highest point of trajectory is zero. Setting $v_y=0$, one can get the time between initial time and where the projectile reaches the highest point. Thus, use the equation for projectile vertical velocity at any time as \begin{align*} v_y&=v_0\,\sin \theta-gt\\ 0&=150\times \sin 37^\circ-(9.8)\,t \end{align*} Solving the above linear equation for $t$, we get the time between firing and highest points as $t=9.2\,{\rm s}$. Substituting it into the projectile formula for vertical displacement, we have \begin{align*} y&=-\frac 12 gt^2 + (v_0\,\sin \theta)\,t+y_0 \\ h&=-\frac 12 (9.8)(9.2)^2 + (150\times \sin 37^\circ)(9.2)+0\\ h&=415.76\,{\rm m} \end{align*} In the second line, $y_0$ is set to zero since from the beginning of problem we adopted the origin of the coordinate $(x_0=0,y_0=0)$ at the firing point. Adding this value with the cliff height, the total height the projectile reaches from the ground is obtained. $H=d+h=200+415.76=615.76\,{\rm m}$

(c) To find the velocity of a projectile at any time, we require to compute its components at any instant of time. First calculate the vertical and horizontal components of velocity and then use the Pythagorean theorem to find the resultant velocity vector as below
\begin{align*} \text{Vertical speed}: \, v_y&=(v_0 \sin \theta)-gt\\ &=(150\times \sin 37^\circ)-(9.8\times 20.757)\\ v_y&=-113.14\,{\rm m/s}\\ \text{Horizontal speed}:\, v_x&=v_0\,\cos \theta \\ &=150\times \cos 37^\circ \\ v_x&=119.79\,{\rm m/s} \end{align*} Note that in above we put the total time in the vertical speed formula. As expected, the vertical component has a minus sign indicates that the projectile direction is down. Given those components, it is easy task to find the magnitude and direction of the velocity vector \begin{align*} v&=\sqrt{v_x^2 + v_y^2}\\
&=\sqrt{(119.79)^2 + (-113.14)^2}\\ &=164.77\,{\rm m/s}\\ \text{AND}\\ \theta &= \tan^{-1} \left(\frac{v_y}{v_x}\right)\\ &=\tan^{-1} \left(\frac{-113.14}{119.79}\right)\\ &=-43.36^\circ \end{align*} The negative indicates angle is below the horizontal axis. Example (2):

A bird carrying a juniper berry suddenly releases the berry when it is $30\,{\rm m}$ above the level ground. At the instant the berry is released it has a velocity of $10\,{\rm m/s}$ at an angle of $20^\circ$ from horizontal. Find:
(a) The time the berry reaches the ground.
(b) The velocity vector of the berry when it reaches the ground.

Solution:

This is a projectile motion problem since the berry has an angle and velocity and under the effect of gravity reaches the ground.
(a) Let the releasing point be the origin of coordinate i.e. $(x_0=0,y_0=0)$. Since the berry hit the ground below the y-axis so the coordinate of impact is $(x=?,y=-h=-30)$, where $h$ is the vertical distance from bird to the hitting point.
\begin{align*} y&=-\frac 12 gt^2 + (v_0\,\sin \theta)\,t+y_0\\ -30&=-\frac 12 (9.8)\,t^2 + (10\times \sin 20^\circ)\,t+0\\ & \Rightarrow \,t_1=2.848\,{\rm s} \ , \ t_2=-2.15\,{\rm s}
\end{align*} Since the berry dropped at the initial time $t=0$, so the total time the berry is in the air (accepted time) is $t_1=2.848\,{\rm s}$.

(b) The components of the velocity vector is determined as \begin{align*}
\text{Vertical component}: \, v_y&=(v_0 \sin \theta)-gt\\ &=(10\times \sin 20^\circ)-(9.8\times 2.848)\\ v_y&=-24.5\,{\rm m/s}\\ \text{Horizontal component}:\, v_x &=v_0\,\cos \theta \\ &=10\times \cos 20^\circ \\ v_x &=9.4\,{\rm m/s}
\end{align*} Thus, the vector addition of those components gets the velocity vector. \begin{align*} \vec{v}&=v_x\,\hat{i} + v_y\,\hat{j}\\ &=9.4\,\hat{i} -24.5\,\hat{j}
\end{align*}

Example (3) for horizontal projectile motion:
A stone is thrown horizontally into the air with a speed of $8\,{\rm m/s}$ from the top of a $20\,{\rm m}$-high cliff and hits the ground.
(a) At what distance from the base of the cliff does the stone land?
(b) Find the speed and direction of the stone just before it hits the ground?

Solution:
Since said that the stone is thrown horizontally, so $\theta=0$. Let the origin of the coordinate be the throwing point that is $(x_0=0,y_0=0)$ and the coordinate of the landing point $(x=?,y=-20\,{\rm m})$.
(a) The horizontal distance from the base is obtained by $x=v_0\,t$, in which $t$ is the time from base to the desire point. The time between throwing to landing points is called total flight time $T$ and is obtained by $\Delta y=-\frac 12\,g\,t^2$. Thus, we get
\begin{align*} \Delta y&=-\frac 12\,g\,t^2 \\ -24&=-\frac 12 \,(9.8)\,T^2\\ &=2.21\,{\rm s}
\end{align*} Now, substitute it into the equation for horizontal distance \begin{align*} x&=(v_0\,\cos \theta)\,t\\ &=(8\times \cos 0)\,(2.21)\\ &=17.68\,{\rm m} \end{align*}

(b) The velocity's components of the projectile just before hitting the ground is
\begin{align*} v_{x}&=v_0=8\,{\rm m/s}\\ v_y&=-gT=-9.8\times 2.21\\ &=-21.7\,{\rm m/s}\\
\text{speed}&=\sqrt{v_x^2 + v_y^2}\\ &=\sqrt{8^2 + (21.7)^2}\\ &=23.12\,{\rm m/s}\\
\text{And}\\ \text{direction}\ \theta &=\tan^{-1}\left(\frac{v_y}{v_x}\right)\\ &=\tan^{-1}\left(\frac{-21.7}{8}\right)\\ &=-69.7^\circ \end{align*} The negative indicates the angle is below the positive $x$-axis.

Example (4):
A pirate ship is $560\,{\rm m}$ away from a fort defending a harbor entrance. A defense cannon at sea level fires balls at an initial velocity of $82\,{\rm m/s}$ at an angle of $63^\circ$. (a) How long is the cannon ball in the air?  (b) What is the maximum height reached by the ball?

Solution: Given data: the maximum distance that the cannon-balls hit the ships i.e. $\Delta x=560\,{\rm m}$, initial velocity $v_0=82\,{\rm m/s}$ and the angle made by the cannon with the horizontal is $63^\circ$.
(a) \begin{align*} \Delta y&=-\frac 12 gt^2 +\left(v_0 \sin \theta \right)t\\ 0&=-\frac 12 (9.8)t_T^2 +\left(82\times \sin 63^\circ\right)t_T \end{align*} Thus, the total time that the cannon balls are in the air is $t_T=14.91\,{\rm s}$.
(b) When the vertical component of projectiles’ velocity is zero i.e. $v_y=0$, the projectile is in the highest point of its parabolic trajectory, so using the following kinematic equation for displacement in the vertical direction we have \begin{align*}
v_y^2-v_{0y}^2  &=2(-g)\Delta y \\ v^2 - \left(v_0 \sin \theta \right)^2 &=2(-g)\Delta y \\
0-(82\times \sin 63^\circ)^2 &=2(-9.8)H\\ \Rightarrow H&=272.3\,{\rm m} \end{align*}

Problem (5):
A soccer ball is kicked at $35^\circ$ above the horizontal, it lands $13.8\,{\rm m}$ away and reaches a maximum height of $2.42\,{\rm m}$.
a) What is the initial vertical velocity?
b) How long does it take to hit the ground?
c) What is the initial horizontal velocity?
d) With what velocity was the ball initially kicked ($\vec{v}$).

Solution:
Known: $\theta=35^\circ$, Range of projectile $R=13.8\,{\rm m}$, Maximum height $H=2.42\,{\rm m}$.

(a) Recall that in the projectile problems there are two motions and consequently two acceleration and velocity.

Motion along the horizontal direction is uniform ($a_x=0$) and in vertical direction is free-fall motion ($a_y=-g$).

Velocity is a vector quantity so its component at the launching point are $v_{0x}=v_0 \cos \theta$ and $v_{0y}=v_0 \sin \theta$.

In this part $v_{0y}$ is unknown. Thus, either first, find the initial velocity of $v_0$ and then its vertical component or use the data for maximum height.

Recall that the projectile range is determined by
$R=\frac{v_0^2 \sin 2\theta}{g}$
After substituting the given data into it, obtain \begin{align*} 13.8 &= \frac{v_0^2 \sin 2\times 35^\circ}{9.8}\\  \Rightarrow v_0 &=11.99\, {\rm m/s} \end{align*} Therefore, \begin{align*} v_{0y}&=v_0 \sin \theta\\ &=(11.99)(\sin 35^\circ)\\ &=6.87\,{\rm m/s} \end{align*} Or using the maximum height, we have \begin{align*} v_y^2 -v_{0y}^2 &=2(-g)\Delta y\\ 0-v_{0y}^2 &=-2(9.8)(2.42)\\ \Rightarrow v_{0y} &= 6.87\,{\rm m/s} \end{align*}
(b) Initial and final points of the projectile are at the same level i.e. $\Delta y=0$, so by setting this into the following kinematic equation along vertical direction, we obtain
\begin{align*} \Delta y &=-\frac 12 gt^2 +\underbrace{v_0 \sin \theta}_{v_{0y}} t\\
0&=-\frac 12 (9.8)t_T^2 +(6.87) t_T \end{align*} Thus, the time of flight is $t_T=1.40\,{\rm s}$.

(c) In projectile problems, horizontal component of initial velocity appears in the uniform motion along the horizontal direction as $\Delta x=v_{0x} t$, where $\Delta x$  is the horizontal distance. When total time of flight is substituted for $t$, then $\Delta x$ equals the range of projectile.

In this part, $v_{0x}$ is requested so
\begin{align*} \Delta x &= R=v_{0x} t_T \\ 13.8 &= v_{0x} (1.40)\\ \Rightarrow v_{0x} &=\frac{13.8}{1.4}=9.85\,{\rm m/s} \end{align*}
(d) With having the vertical and horizontal components of projectile’s velocity, we can find the resultant velocity vector as bellow
\begin{align*} \vec{v}&=\sqrt{v_{0x}^2 +v_{0y}^2}\\ &=\sqrt{(9.85)^2+(6.87)^2}\\     &\approx 12\,{\rm m/s} \end{align*} Note: the horizontal component of projectile's velocity is always the same throughout the projectile path.

Practice more problems - Kinematics in Two Dimensions.