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Motion in Two Dimensions Questions

Questions

A small block is placed at height  $h$ on a frictionless,  $30{}^\circ $ ramp.  Upon  being released (from rest), the  block  slides  down the  ramp  and  then falls  $1.0\ \mathrm{m}$ to  the  floor.  A small hole is located $1.0\ \mathrm{m}$ from the end of the ramp.  From what  height $h$ should the  block  be released  in  order to land  in  the  hole?

First, find the speed of the block at the bottom of ramp (from conservation of energy):
\[U_i+K_i=U_f+K_f\to \ \ mgh=\frac{1}{2}mv^2_1\to v_1=\sqrt{2gh}\] 
In the end of the ramp, block moving in two direction (projectile in plane):
The horizontal distance that the block hits the ground to the base is given by $\mathrm{\Delta }x=v_{1x}t+\frac{1}{2}a_xt^2$ but in $x$ direction there is a uniform motion i.e. $a_x=0$, so $t=\frac{\mathrm{\Delta }x}{v_{1x}}=\frac{\mathrm{\Delta }x}{v_1\,{\cos \theta \ }}$ .We know that in the projectiles motion the time traveled in horizontal direction is equal to that of in vertical direction. Then,
\begin{align*}
\Delta y &=v_{1y}t+\frac{1}{2}a_yt^2\\
&=v_1\,{\sin \theta \ }\frac{\mathrm{\Delta }x}{v_1\,{\cos \theta \ }}+\frac{1}{2}\left(-g\right){\left(\frac{\mathrm{\Delta }x}{v_1{\cos \theta \ }}\right)}^2
\end{align*} 
\[\mathrm{\Rightarrow }\mathrm{\ \ \ }\mathrm{\Delta }y=\mathrm{\Delta }x\,{\tan \theta \ }-\frac{g{\left(\mathrm{\Delta }x\right)}^2}{2v^2_1{{\cos}^{\mathrm{2}} \theta \ }}\] 
Solving for $v_1$, we get 
\begin{gather*}
\frac{g{\left(\mathrm{\Delta }x\right)}^2}{2v^2_1{{\cos}^{\mathrm{2}} \theta \ }}=\mathrm{\Delta }x\,{\tan \theta \ }-\mathrm{\Delta }y\\
\Rightarrow \ v^2_1=\frac{g{\left(\mathrm{\Delta }x\right)}^2}{2{{\cos}^2 \theta \ }\left(-\mathrm{\Delta }y+\mathrm{\Delta }x{\tan \theta \ }\right)}=2gh
\end{gather*} 
Then the desired height is obtained as
\[\therefore h=\frac{{\left(\mathrm{\Delta }x\right)}^2}{4{{\cos}^2 \theta \ }(-\mathrm{\Delta }y+\mathrm{\Delta }x\,{\tan \theta \ })}\] 
Consider the end of the ramp as the origin of coordinate system, so the coordinate of the landing point is $(x=1\mathrm{m,y=-1m)}$ and since the block exit the ramp at $30{}^\circ $ below the horizontal so $\theta =-30{}^\circ $. Substituting these values into the above relation for $h$, we obtain
\[h=\frac{{\left(1\right)}^2}{4{{\cos}^2 \left(-30{}^\circ \right)\ }\left(-\left(-1\right)+1{\tan \left(-30{}^\circ \right)\ }\right)}=0.788\ \mathrm{m}\] 
 

A missile is shot horizontally from the top of a $500\ \mathrm{m}$ cliff with an initial speed of $300\ \mathrm{m/s}$
(a) Find the time it takes for the missile to hit the ground.
(b) What is the range of the missile?
(c) Calculate the velocity of the missile just before it hits the ground.
(d) If the missile hits the ground and bounces up at an angle of $30{}^\circ$ with a speed of $\mathrm{200\ m/s}$, how far away from the point of impact will it land?
 

Projectile motion: in this motion without air resistance we have $a_x=0$ and $a_y=-g$. The kinematic relations for projectile are:
\begin{gather*}
x=\underbrace{\left(v_0\,{\cos \alpha \ }\right)}_{v_{0x}}t\\
y=-\frac{1}{2}gt^2+\underbrace{\left(v_0\,{\sin \alpha \ }\right)}_{v_{0y}}t+y_0
v_x=\underbrace{v_0\,{\cos \alpha \ }}_{v_{0x}}\ \ ,\ \ v_y=\underbrace{v_0\,{\sin \alpha \ }}_{v_{0y}}-gt
\end{gather*}
(a) If we choose the initial position of the missile as the origin of the coordinate system (i.e. $x_0=y_0=0$)then the hitting position has coordinate ($x,y=-500\,\mathrm{m}$). Use the following kinematic relation to find the total flight time. 
\begin{gather*}
y=-\frac{1}{2}gt^2+v_0t\,{\sin \alpha \ }+y_0\\
\to \ -500=-\frac{1}{2}\left(9.8\right)t^2+300\,t\,{\sin 0{}^\circ \ }+0
\end{gather*}
\[\Rightarrow t_{tot}=\sqrt{\frac{1000}{9.8}}=10.10\ \mathrm{s}\] 
Note: in horizontally shot $\alpha =0{}^\circ $
(b) To find the range of the projectile, we must find the total flight time of the motion then substitute it into $x=v_0t\,{\cos \alpha \ }$. 
In the previous part the total flight time is calculated as $t=10.10\,\mathrm{s}$, therefore 
\begin{align*}
R_1=x&=v_0t\,{\cos \alpha \ }\\
&=300\times 10.10\times {\cos 0{}^\circ \ }\\
&=3000.30\ \mathrm{m}
\end{align*}
(c) In projectile motion first find the components of the velocity then use the $v=\sqrt{v^2_x+v^2_y}$ to determine the velocity of the missile at any moment (in this case the hitting position, $t_{tot}$). 
\[v_x=v_0\,{\cos \alpha \ }=300\times {\cos 0{}^\circ \ }=300\frac{\mathrm{m}}{\mathrm{s}}\] 
\begin{align*}
v_y&=v_0\,{\sin \alpha \ }-gt\\
&=300\times {\sin 0{}^\circ \ }-\left(9.8\times 10.10\right)\\
&=-98.98\frac{\mathrm{m}}{\mathrm{s}}
\end{align*}
\[v=\sqrt{v^2_x+v^2_y}=\sqrt{{\left(300\right)}^2+{\left(-98.98\right)}^2}=315.9\ \mathrm{m/s}\] 
(d) In this part we have a separate projectile motion problem, hence we choose the launching point as the origin with the following information $x_0=y_0=0\ ,\ v_0=200\ \mathrm{m/s}$ , $\alpha =30{}^\circ $
. First by setting $y=0$ in the $y=-\frac{1}{2}gt^2+v_0t\,{\sin 30{}^\circ \ }+y_0$ find the total fight time, then substitute it into the $x=v_0t\,{\cos 30{}^\circ \ }$
\[0=-\frac{1}{2}\left(9.8\right)t^2+200\,t\left(\frac{1}{2}\right)+0\Rightarrow t\left(9.8t-200\right)=0\] 
\[\Rightarrow \left\{ \begin{array}{rcl}
t_1 & = & 0\ ,\ \ \ \mathrm{initial\ time} \\ 
t_2 & = & \frac{200}{9.8}=20.4\ \mathrm{s\ \ ,\ \ \ landing\ time} \end{array}
\right.\] 
\begin{align*}
R_2=x &=v_0t_2\,{\cos 30{}^\circ \ }\\
&=200\times 20.4\times \left(\frac{\sqrt{3}}{2}\right)\\
&=3533.38\ \mathrm{m}
\end{align*}


 

A baseball is thrown from the ground into the air with an initial velocity $\vec{v}=20\ \hat{i}+10\ \hat{j}\ (\frac{\mathrm{m}}{\mathrm{s}}\mathrm{)}$
(a) To what maximum height will the ball rise?
(b) What will be the total horizontal distance travelled by the ball?
(c) What is the velocity of the ball the instant before it hits the ground?
 

(a) From the given velocity we can determine the launch angle and the magnitude of initial velocity.
\begin{gather*}
\alpha ={{\tan}^{-1} \frac{v_y}{y_x}\ }={{\tan}^{-1} \left(\frac{10}{20}\right)\ }=26.5{}^\circ\\
v_0=\sqrt{{10}^2+{20}^2}=22.36\ \frac{\mathrm{m}}{\mathrm{s}}
\end{gather*}
To find the maximum height we must calculate when the vertical component of velocity is zero $v_y=0$ then substitute it into the vertical distance relation.
\begin{gather*}
v_y=v_0\,{\sin 26.5{}^\circ \ }-gt=0\\
\Rightarrow t=\frac{v_0\,{\sin 26.5{}^\circ \ }}{g}=\frac{22.36\times 0.44}{9.8}=1.01\ \mathrm{s}
\end{gather*}
\begin{align*}
y&=-\frac{1}{2}gt^2+v_0t\,{\sin \alpha \ }+y_0\\
&=-\frac{1}{2}\left(9.8\right){\left(1.01\right)}^2+22.36\left(1.01\right){\sin 26.5{}^\circ \ }+0\\
&=5.07\ \mathrm{m}
\end{align*}
Alternative solution: without finding the angle and magnitude of the velocity we can find the maximum height!
\begin{gather*}
\vec{v}=\underbrace{20}_{v_{0x}}\hat{i}+\underbrace{10}_{v_{0y}}\hat{\mathrm{j}}\ \left(\frac{\mathrm{m}}{\mathrm{s}}\right)\\
v_y=v_{0y}-gt=0\Rightarrow t=\frac{v_{0y}}{g}=\frac{10}{9.8}=1.01\ \mathrm{s}
\end{gather*}
\begin{align*}
y&=-\frac{1}{2}gt^2+\underbrace{v_0\,{\sin \alpha \ }}_{v_{0y}}t+y_0\\
&=-\frac{1}{2}\left(9.8\right){\left(1.01\right)}^2+10\left(1.01\right)+0\\
&=5.10\ \mathrm{m}
\end{align*}
Small difference between the two results is due to uncertainty in the magnitude of the angle and initial velocity of the projectile. 
(b) To find the horizontal distance, calculate the total flight time and then put it in the $x$ relation.
For finding the total flight time, substitute the location of hitting the projectile to the ground into $y=-\frac{1}{2}gt^2+v_{0y}t+y_0$. Because the launching and landing point are in the same level (choose this point as the origin $x_0=y_0=0$) set $y=0$.
\[0=-\frac{1}{2}\left(9.8\right)t^2_{tot}+10\,t_{tot}+0\Rightarrow t_{tot}\left(9.8\ t_{tot}-20\right)=0\] 
\[\Rightarrow t_{tot}=\left\{ \begin{array}{rcl}
0\ & , & \ \ \mathrm{initial\ time} \\ 
\frac{20}{9.8} & = & 2.04\ \mathrm{s\ ,\ \ landing\ time} \end{array}
\right.\] 
\begin{gather*}
x=v_0\, {\cos \alpha \ }t=v_{0x}t\\
\Rightarrow R=x_{tot}=v_{0x}t_{tot}=20\times 2.04=40.8\ \mathrm{m}
\end{gather*}
This is the maximum distance traveled in the horizontal direction or range of the projectile.
(c) First substitute the total flight time into the $v_y$ and $v_x$ then find the magnitude of the velocity.
\[\left\{ \begin{array}{c}
v_x=v_{0x}=20\frac{\mathrm{m}}{\mathrm{s}} \\ 
v_y=v_{0y}-gt_{tot}=10-\left(9.8\right)\left(2.04\right)=-9.992\frac{\mathrm{m}}{\mathrm{s}} \end{array}
\right.\] 
\[\Rightarrow v=\sqrt{v^2_x+v^2_y}=\sqrt{{20}^2+{\left(-9.992\right)}^2}=22.35\frac{\mathrm{m}}{\mathrm{s}}\] 
 

A projectile is fired horizontally from a gun that is $45\ \mathrm{m}$ above the ground. The muzzle velocity is $250\ \mathrm{m/s}$. 
(a) How long does the projectile remains in air?
(b) At what horizontal distance from the firing point does it strike the ground?
(c) What is the magnitude of the vertical component of its velocity as it strikes the ground?
 

(a) Remains in air means that the total flight time. To find this time put the coordinate of the impact of the projectile into the $y=-\frac{1}{2}gt^2+v_0t\,{\sin \alpha \ }+y_0$ and then solve for the time $t_{tot}$. Let the origin of the coordinates to be the firing point. Therefore, the projectile hits the ground $-45\ \mathrm{m}$below the origin!
\[y=-\frac{1}{2}gt^2+v_0\,{\sin \alpha \ }\ t+y_0\] 
\[-45=-\frac{1}{2}\left(9.8\right)t^2+250\,t\,{\sin 0{}^\circ \ }+0\]
\[\Rightarrow t_{tot}=\sqrt{\frac{90}{9.8}}=3.03\mathrm{s}\] 
(b) Namely, find the range of the projectile. Hence substitute $t_{tot}$ into the $x$ component of the motion i.e.
\[R=x=v_0\,{\cos \alpha \ }t_{tot}=\left(250\right)\left(3.03\right){\cos 0{}^\circ \ }=757.5\ \mathrm{m}\] 
(c) The components of the velocity in a projectile motion at any moment in time are
\[v_x=v_{0x}=v_0\,{\cos \alpha \ }=250\,{\cos 0{}^\circ \ }=250\frac{\mathrm{m}}{\mathrm{s}}\] 
\begin{align*}
v_y&=v_{0y}-gt=v_0\,{\sin \alpha \ }-gt\\
&=250\,{\sin 0{}^\circ \ }-\left(9.8\right)\left(3.03\right)\\
&=29.69\ \frac{\mathrm{m}}{\mathrm{s}}
\end{align*} 
Where we have substitute the total time (the moment of impact on the ground).
 

A ball is thrown horizontally from the roof of a building $50\ \mathrm{m}$ tall and lands $45\,\mathrm{m}$ from the base. What was the ball's initial speed?

This is a projectile motion with launch angle $\alpha =0{}^\circ $, so use the projectile equations as 
\begin{gather*}
x=v_{0x}t=v_0\,{\cos \alpha \ }t\ \ ,\ \ y=-\frac{1}{2}gt^2+\underbrace{v_0{\sin \alpha \ }}_{v_{0y}}t+y_0\\
v_x=v_0\,{\cos \alpha \ }\ \ ,\ \ v_y=\underbrace{v_0\,{\sin \alpha \ }}_{v_{0y}}-gt
\end{gather*}
If we choose the releasing point as the reference then the coordinate of the point of impact is $(x=45,y=-50\ \mathrm{m)}$. First find the total flight time then substitute it into the $x$ component of the projectile.
\begin{gather*}
y=-\frac{1}{2}gt^2_{tot}+\underbrace{v_0{\sin \alpha \ }}_{v_{0y}}t_{tot}+y_0\\
\Rightarrow \ -50=-\frac{1}{2}\left(9.8\right)t^2_{tot}+v_0\,{\sin 0{}^\circ \ }t_{tot}+0\\
\Rightarrow t_{tot}=\sqrt{\frac{2\times 50}{9.8}}=3.19\ \mathrm{s\ }
\end{gather*}
\[x=v_0\,{\cos \alpha \ }t_{tot}\Rightarrow v_0=\frac{x}{{\cos 0{}^\circ \ }\,t_{tot}}=\frac{45}{3.19}=14.1\frac{\mathrm{m}}{\mathrm{s}}\] 


 

A $1\ \mathrm{kg}$ projectile is fired from a cannon with an initial kinetic energy of ${10}^4\ \mathrm{J}$. The cannon has an elevation angle of $45{}^\circ $. How far does the projectile go before striking the ground (neglect air resistance)? 

In the projectile language, the distance from the launching to striking point is called the range of the projectile which is found by substituting the total flight time into the $x$ component of the projectile motion that is $X=v_0\,{\cos \theta \ }\ t$. 
From the definition of the kinetic energy one can find the initial velocity of the projectile.
\[K=\frac{1}{2}mv^2_0\to {10}^4=\frac{1}{2}\left(1\right)v^2_0\] 
\[\Rightarrow v_0=\sqrt{2}\times {10}^2\frac{\mathrm{m}}{\mathrm{s}}\] 
Consider the starting and landing points to be in the same level. In this case, using the kinematic equation $v_y=v_0\,{\sin \theta \ }-gt$ and knowing the fact that at the highest point the vertical component of the projectile's velocity is zero i.e. $v_y=0$, find half of the total flight time that is $t_{tot}=2t$ (since there is no air resistance).
\begin{gather*}
v_y=v_0\,{\sin \theta \ }-gt=0\\
\to t=\frac{v_0\,{\sin \theta \ }}{g}=\frac{\left(\sqrt{2}\times {10}^2\right){\sin 45{}^\circ \ }}{9.8}=10.2\ \mathrm{s}
\end{gather*} 
This is the elapsed time to the highest point.
\[\Rightarrow t_{tot}=2t=20.4\ \mathrm{s}\] 
Therefore, 
\begin{align*}
Range=X&=v_0t\,{\cos \theta \ }\\
&=\left(\sqrt{2}\times {10}^2\right){\cos 45{}^\circ \ }\times 20.4\\
&=2040\ \mathrm{m}
\end{align*}
 


Motion in Two Dimensions
Category : Motion in Two Dimensions

MOST USEFUL FORMULA IN MOTION IN TWO DIMENSION:

Position vector $\vec r$ points from the origin of the coordinate system to the particle.
\[\vec r=x\hat i+y\hat j+z\hat k\]
Average velocity:
\[\vec v_{ave}=\frac{\vec r_2-\vec r_1}{t_2-t_1}=\frac{\Delta \vec r}{\Delta t}\]
Instantaneous velocity:
\[\vec v=\lim_{\Delta t \to 0} \frac {\Delta \vec r}{\Delta t}=\frac{d\vec r}{dt}\]
Average acceleration:
\[\vec a_{ave}=\frac{\vec v_2-\vec v_1}{t_2-t_1}=\frac{\Delta \vec v}{\Delta t}\]
Instantaneous acceleration:
\[\vec a=\lim_{\Delta t \to 0} \frac {\Delta \vec v}{\Delta t}=\frac{d\vec v}{dt}\]

Projectile motion:
Acceleration:
\[a_x=0 \quad , \quad a_y=-g\]
Coordinate components:
\[x=\left(v_0\,\cos \alpha\right)t \quad , \quad y=\left(v_0\,\sin \alpha \right)t-\frac 1 2 gt^2\]
velocity components:
\[v_x=v_0\,\cos \alpha\ \quad , \quad v_y=v_0\,\sin \alpha -gt\]


Number Of Questions : 6