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AP Physics 1 : Linear Motion and Momentum

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Example Questions

Example Question #47 : Motion In Two Dimensions

A car traveling at $100\ \frac{\textup{m}}{\textup{s}}$ $\displaystyle 100\ \frac{\textup{m}}{\textup{s}}$ launches off a ramp of height $2\textup{ m}$ $\displaystyle 2\textup{ m}$ and length $5\textup{ m}$ $\displaystyle 5\textup{ m}$. Determine the "hang time" of the car. Ignore air resistance.

Possible Answers:

$3.95\textup{ s}$ $\displaystyle 3.95\textup{ s}$

$6.11\textup{ s}$ $\displaystyle 6.11\textup{ s}$

$7.61\textup{ s}$ $\displaystyle 7.61\textup{ s}$

$8.44\textup{ s}$ $\displaystyle 8.44\textup{ s}$

None of these

Correct answer:

$7.61\textup{ s}$ $\displaystyle 7.61\textup{ s}$

Explanation:

Finding the angle at which the car launches at:

y:2 $\displaystyle y:2$

x:5 $\displaystyle x:5$

$\theta=tan^{-1}(\frac{2}{5})$ $\displaystyle \theta=tan^{-1}(\frac{2}{5})$

$\theta=21.8^\circ$ $\displaystyle \theta=21.8^\circ$

Determining $\displaystyle y$ component of velocity:

$100\frac{m}{s}*sin(21.8^\circ)=37.1\frac{m}{s}$ $\displaystyle 100\frac{m}{s}*sin(21.8^\circ)=37.1\frac{m}{s}$

Using conservation of energy in a closed system:

E_i=E_f $\displaystyle E_i=E_f$

Initially, the car has both kinetic and gravitational potential energy in the y direction, but after, it only has gravitational potential.

$\displaystyle .5mv_y^2+mgh_i=mgh_f$

Solving for h_f $\displaystyle h_f$

$\frac{v_y^2+gh_i}{g}=h_f$ $\displaystyle \frac{v_y^2+gh_i}{g}=h_f$

Plugging in values:

$\frac{37.1^2+9.8*2}{9.8}=h_f$ $\displaystyle \frac{37.1^2+9.8*2}{9.8}=h_f$

$\displaystyle h_f=142.45m$

Determine the velocity in the y direction when the car hits the ground

$\displaystyle .5mv_f^2=mgh_i$

Solving for v_f $\displaystyle v_f$

$v_f=\sqrt{gh_i}$ $\displaystyle v_f=\sqrt{gh_i}$

Plugging in values:

$v_f=\sqrt{9.8*142.45}$ $\displaystyle v_f=\sqrt{9.8*142.45}$

$v_f=37.4\frac{m}{s}$ $\displaystyle v_f=37.4\frac{m}{s}$

Now, breaking the jump into two parts, the "up" and the "down" and using $\displaystyle v_i+at=v_f$

Up:

$37.1\frac{m}{s}-9.8*t_{up}=0$ $\displaystyle 37.1\frac{m}{s}-9.8*t_{up}=0$

Down:

$0\frac{m}{s}+9.8*t_{down}=37.4\frac{m}{s}$ $\displaystyle 0\frac{m}{s}+9.8*t_{down}=37.4\frac{m}{s}$

$t_{up}=3.79s$ $\displaystyle t_{up}=3.79s$

$t_{down}=3.82s$ $\displaystyle t_{down}=3.82s$

$t_{up}+t_{down}=7.61s$ $\displaystyle t_{up}+t_{down}=7.61s$

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Example Question #48 : Motion In Two Dimensions

Possible Answers:

None of these

$740\textup{ m}$ $\displaystyle 740\textup{ m}$

$707\textup{ m}$ $\displaystyle 707\textup{ m}$

$626\textup{ m}$ $\displaystyle 626\textup{ m}$

$819\textup{ m}$ $\displaystyle 819\textup{ m}$

Correct answer:

$707\textup{ m}$ $\displaystyle 707\textup{ m}$

Explanation:

Finding the angle at which the car launches at:

y:2 $\displaystyle y:2$

x:5 $\displaystyle x:5$

$\theta=tan^{-1}(\frac{2}{5})$ $\displaystyle \theta=tan^{-1}(\frac{2}{5})$

$\theta=21.8^\circ$ $\displaystyle \theta=21.8^\circ$

Determining $\displaystyle y$ component of velocity:

$100\frac{m}{s}*sin(21.8^\circ)=37.1\frac{m}{s}$ $\displaystyle 100\frac{m}{s}*sin(21.8^\circ)=37.1\frac{m}{s}$

Using conservation of energy in a closed system:

E_i=E_f $\displaystyle E_i=E_f$

Initially, the car has both kinetic and gravitational potential energy in the y direction, but after, it only has gravitational potential.

$\displaystyle .5mv_y^2+mgh_i=mgh_f$

Solving for h_f $\displaystyle h_f$

$\frac{v_y^2+gh_i}{g}=h_f$ $\displaystyle \frac{v_y^2+gh_i}{g}=h_f$

Plugging in values:

$\frac{37.1^2+9.8*2}{9.8}=h_f$ $\displaystyle \frac{37.1^2+9.8*2}{9.8}=h_f$

$\displaystyle h_f=142.45m$

Determine the velocity in the y direction when the car hits the ground

$\displaystyle .5mv_f^2=mgh_i$

Solving for v_f $\displaystyle v_f$

$v_f=\sqrt{gh_i}$ $\displaystyle v_f=\sqrt{gh_i}$

Plugging in values:

$v_f=\sqrt{9.8*142.45}$ $\displaystyle v_f=\sqrt{9.8*142.45}$

$v_f=37.4\frac{m}{s}$ $\displaystyle v_f=37.4\frac{m}{s}$

Now, breaking the jump into two parts, the "up" and the "down" and using $\displaystyle v_i+at=v_f$

Up:

$37.1\frac{m}{s}-9.8*t_{up}=0$ $\displaystyle 37.1\frac{m}{s}-9.8*t_{up}=0$

Down:

$0\frac{m}{s}+9.8*t_{down}=37.4\frac{m}{s}$ $\displaystyle 0\frac{m}{s}+9.8*t_{down}=37.4\frac{m}{s}$

$t_{up}=3.79s$ $\displaystyle t_{up}=3.79s$

$t_{down}=3.82s$ $\displaystyle t_{down}=3.82s$

$t_{up}+t_{down}=7.61s$ $\displaystyle t_{up}+t_{down}=7.61s$

Multiplying by the horizontal velocity:

$7.61s*100\frac{m}{s}cos(21.8^\circ)=707m$ $\displaystyle 7.61s*100\frac{m}{s}cos(21.8^\circ)=707m$

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Example Question #141 : Linear Motion And Momentum

Deep in space Object $\displaystyle A$ has mass 350kg $\displaystyle 350kg$ and is initially traveling with velocity $<30,50>\frac{m}{s}$ $\displaystyle < 30,50>\frac{m}{s}$. At t=0 $\displaystyle t=0$, it collides with Object $\displaystyle B$, which has mass 500kg $\displaystyle 500kg$ and is initially motionless. The two objects stick together.

Assuming the collision took place at the origin, determine the location at t=55s $\displaystyle t=55s$, fifty five seconds after the collision.

Possible Answers:

<682,1133>m $\displaystyle < 682,1133>m$

None of these

<545,1138>m $\displaystyle < 545,1138>m$

<626,395>m $\displaystyle < 626,395>m$

<808,225>m $\displaystyle < 808,225>m$

Correct answer:

<682,1133>m $\displaystyle < 682,1133>m$

Explanation:

Using

$\overrightarrow{p}=m\overrightarrow{v}$ $\displaystyle \overrightarrow{p}=m\overrightarrow{v}$

Plugging in values:

$\overrightarrow{p}=350kg*<30,50>\frac{m}{s}$ $\displaystyle \overrightarrow{p}=350kg*< 30,50>\frac{m}{s}$

$\overrightarrow{p}=<10500,17500>\frac{kg*m}{s}$ $\displaystyle \overrightarrow{p}=< 10500,17500>\frac{kg*m}{s}$

The momentum will be the same in the final state, so again using

$\overrightarrow{p}=m\overrightarrow{v}$ $\displaystyle \overrightarrow{p}=m\overrightarrow{v}$

Solving for velocity:

$\frac{\overrightarrow{p}}{m}=\overrightarrow{v}$ $\displaystyle \frac{\overrightarrow{p}}{m}=\overrightarrow{v}$

Plugging in values (the total mass is equal to the combined masses:

$\frac{<10500,17500>}{850}=\overrightarrow{v}$ $\displaystyle \frac{< 10500,17500>}{850}=\overrightarrow{v}$

$\overrightarrow{v_f}=<12.4,20.6>\frac{m}{s}$ $\displaystyle \overrightarrow{v_f}=< 12.4,20.6>\frac{m}{s}$

Using

$r_i+\overrightarrow{v}*\Delta t=r_f$ $\displaystyle r_i+\overrightarrow{v}*\Delta t=r_f$

Plugging in values

(0,0)+<12.4,20.6>*55=r_f $\displaystyle (0,0)+< 12.4,20.6>*55=r_f$

r_f=<682,1133>m $\displaystyle r_f=< 682,1133>m$

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Example Question #411 : Ap Physics 1

Assuming the collision took place at the origin, determine the location of object $\displaystyle A$ at t=-15s $\displaystyle t=-15s$, fifteen seconds before the collision.

Possible Answers:

<-450,-750>m $\displaystyle < -450,-750>m$

<450,750>m $\displaystyle < 450,750>m$

<-720,-360>m $\displaystyle < -720,-360>m$

None of these

<-456,-113>m $\displaystyle < -456,-113>m$

Correct answer:

<-450,-750>m $\displaystyle < -450,-750>m$

Explanation:

Using

$r_i+\overrightarrow{v}*\Delta t=r_f$ $\displaystyle r_i+\overrightarrow{v}*\Delta t=r_f$

Plugging in values

r_i+<30,50>*15=(0,0) $\displaystyle r_i+< 30,50>*15=(0,0)$

r_i=<-450,-750>m $\displaystyle r_i=< -450,-750>m$

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Example Question #51 : Motion In Two Dimensions

A ball rolls off of a 200m $\displaystyle 200m$ high cliff at a constant velocity of $20\frac{m}{s}$ $\displaystyle 20\frac{m}{s}$. Neglecting air friction, what is the acceleration of the ball in the vertical and horizontal direction just before it hits the ground?

Possible Answers:

Vertical: $0\frac{m}{s^{2}}$ $\displaystyle 0\frac{m}{s^{2}}$

Horizontal: $0\frac{m}{s^{2}}$ $\displaystyle 0\frac{m}{s^{2}}$

Vertical: $9.8\frac{m}{s^{2}}$ $\displaystyle 9.8\frac{m}{s^{2}}$

Horizontal: $9.8\frac{m}{s^{2}}$ $\displaystyle 9.8\frac{m}{s^{2}}$

Vertical: $-9.8\frac{m}{s^{2}}$ $\displaystyle -9.8\frac{m}{s^{2}}$

Horizontal: $0\frac{m}{s^{2}}$ $\displaystyle 0\frac{m}{s^{2}}$

Vertical: $0\frac{m}{s^{2}}$ $\displaystyle 0\frac{m}{s^{2}}$

Horizontal: $-9.8\frac{m}{s^{2}}$ $\displaystyle -9.8\frac{m}{s^{2}}$

Correct answer:

Vertical: $-9.8\frac{m}{s^{2}}$ $\displaystyle -9.8\frac{m}{s^{2}}$

Horizontal: $0\frac{m}{s^{2}}$ $\displaystyle 0\frac{m}{s^{2}}$

Explanation:

The acceleration due to gravity only affects the object in the vertical direction, which is always $-9.8\frac{m}{s^{2}}$ $\displaystyle -9.8\frac{m}{s^{2}}$. Because there is no external force acting on the ball in the horizontal direction there is no acceleration.

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Example Question #53 : Motion In Two Dimensions

During time period $\displaystyle A$, a rocket ship deep in space of mass $2.55*10^5\textup{ kg}$ $\displaystyle 2.55*10^5\textup{ kg}$ travels from $<0,10>\textup{ km}$ $\displaystyle < 0,10>\textup{ km}$ to $<5,5>\textup{ km}$ $\displaystyle < 5,5>\textup{ km}$. During time period $\displaystyle B$, the rocket fires. During time period $\displaystyle C$, the rocket travels from $<-15,0>\textup{ km}$ $\displaystyle < -15,0>\textup{ km}$ to $<-30,-5>\textup{ km}$ $\displaystyle < -30,-5>\textup{ km}$.

Time periods $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ took $10\textup{ s}$ $\displaystyle 10\textup{ s}$ each.

Determine the momentum during time period $\displaystyle A$.

Possible Answers:

None of these

$<3.985*10^8,-3.985*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$ $\displaystyle < 3.985*10^8,-3.985*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$<1.275*10^8,1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$ $\displaystyle < 1.275*10^8,1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$<-1.275*10^8,1.275*10^8>\frac{\textup{kg} \cdot \textup{m}}{\textup{s}}$ $\displaystyle < -1.275*10^8,1.275*10^8>\frac{\textup{kg} \cdot \textup{m}}{\textup{s}}$

$<1.275*10^8,-1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$ $\displaystyle < 1.275*10^8,-1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Correct answer:

$<1.275*10^8,-1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$ $\displaystyle < 1.275*10^8,-1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Explanation:

$\overrightarrow{P}=m\overrightarrow{v}$ $\displaystyle \overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$ $\displaystyle \overrightarrow{v}=\frac{< x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$ $\displaystyle \overrightarrow{P}=m\frac{< x_f-x_i,y_f-y_i>}{t}$

Plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<5000-0,5000-10000>}{10}$ $\displaystyle \overrightarrow{P}=2.55*10^5*\frac{< 5000-0,5000-10000>}{10}$

$\overrightarrow{P}=<1.275*10^8,-1.275*10^8>\frac{kg*m}{s}$ $\displaystyle \overrightarrow{P}=< 1.275*10^8,-1.275*10^8>\frac{kg*m}{s}$

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Example Question #51 : Motion In Two Dimensions

During time period $\displaystyle A$, a rocket ship deep in space of mass $\displaystyle 2.55*10^5kg$ travels from <0,10>km $\displaystyle < 0,10>km$ to <5,5>km $\displaystyle < 5,5>km$. During time period $\displaystyle B$, the rocket fires. During time period $\displaystyle C$, the rocket travels from <-15,0>km $\displaystyle < -15,0>km$ to <-30,-5>km $\displaystyle < -30,-5>km$.

Time periods $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ all took $\displaystyle 10seconds.$

Determine the momentum during time period $\displaystyle C$.

Possible Answers:

$\overrightarrow{P}=<-3.825*10^8,-1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$ $\displaystyle \overrightarrow{P}=< -3.825*10^8,-1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

None of these

$<3.825*10^8,1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$ $\displaystyle < 3.825*10^8,1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$<3.985*10^8,3.985*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$ $\displaystyle < 3.985*10^8,3.985*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$<8.265*10^8,7.600*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$ $\displaystyle < 8.265*10^8,7.600*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Correct answer:

Explanation:

$\overrightarrow{P}=m\overrightarrow{v}$ $\displaystyle \overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$ $\displaystyle \overrightarrow{v}=\frac{< x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$ $\displaystyle \overrightarrow{P}=m\frac{< x_f-x_i,y_f-y_i>}{t}$

Converting $\displaystyle km$ to $\displaystyle m$ and plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<-30000+15000,-5000-0>}{10}$ $\displaystyle \overrightarrow{P}=2.55*10^5*\frac{< -30000+15000,-5000-0>}{10}$

$\overrightarrow{P}=<-3.825*10^8,-1.275*10^8>\frac{kg*m}{s}$ $\displaystyle \overrightarrow{P}=< -3.825*10^8,-1.275*10^8>\frac{kg*m}{s}$

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Example Question #54 : Motion In Two Dimensions

Suppose that a person has walked $24\: m$ $\displaystyle 24\: m$ north and $12\: m$ $\displaystyle 12\: m$ east. What is this person's net displacement from where he initially started?

Possible Answers:

$36.0\: m$ $\displaystyle 36.0\: m$

$18.0\:m$ $\displaystyle 18.0\:m$

$3.46\:m$ $\displaystyle 3.46\:m$

$26.8\:m$ $\displaystyle 26.8\:m$

$12.0\:m$ $\displaystyle 12.0\:m$

Correct answer:

$26.8\:m$ $\displaystyle 26.8\:m$

Explanation:

To answer this question, it's helpful if we draw the path this person takes.

Vt physics 11 23 15 vector chart

Now, it's important to realize the difference between distance and displacement. The distance this person has traveled is the total amount of walking he has done. So in this case, the distance traveled would be:

$24\:m+12\:m=36\:m$ $\displaystyle 24\:m+12\:m=36\:m$

However, this is not the answer to our question. The question is asking for the displacement, which is the net change in position of this person. We can draw it as follows:

Vt physics 11 23 15 vector displacement

Notice that in the above drawing, we have a right triangle. This will actually simplify things a bit, because we can use trigonometry to find the angle of displacement (shown above in blue). Then, we can use that information to find the displacement (shown above in green).

To find the angle of displacement, we can use the following expression:

$tan(\Theta )=\frac{opposite}{adjacent}=\frac{12\: m}{24\: m}=0.5$ $\displaystyle tan(\Theta )=\frac{opposite}{adjacent}=\frac{12\: m}{24\: m}=0.5$

$\Theta=tan^{-1}(0.5)=26.565^{o}$ $\displaystyle \Theta=tan^{-1}(0.5)=26.565^{o}$

Now that we have the angle, we can calculate the displacement using further trigonometry.

$sin(\Theta )=\frac{opposite}{hypotenuse}=\frac{12\: m}{Displacement}$ $\displaystyle sin(\Theta )=\frac{opposite}{hypotenuse}=\frac{12\: m}{Displacement}$

$sin(26.565)=\frac{12\: m}{Displacement}$ $\displaystyle sin(26.565)=\frac{12\: m}{Displacement}$

$Displacement=\frac{12\: m}{sin(26.565^{o})}=\frac{12\: m}{0.447}=26.8\:m$ $\displaystyle Displacement=\frac{12\: m}{sin(26.565^{o})}=\frac{12\: m}{0.447}=26.8\:m$

An alternative way to solve this problem, and arguably much easier, is to just use the Pythagorean theorem since we have a right triangle.

$Displacement=\sqrt{(12\: m)^{2}+(24\: m)^{2}}=26.8\:m$ $\displaystyle Displacement=\sqrt{(12\: m)^{2}+(24\: m)^{2}}=26.8\:m$

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Example Question #57 : Motion In Two Dimensions

A quarterback throws a football a horizontal distance of 50 yd $\displaystyle 50 yd$ to a wide receiver. The ball was airborne for 5 s $\displaystyle 5 s$. The ball had an initial speed of 50 mph $\displaystyle 50 mph$. The ball has a mass of .43kg $\displaystyle .43kg$.

$1mph=.447\frac{m}{s}$ $\displaystyle 1mph=.447\frac{m}{s}$

$\displaystyle 1 yd=.914 m$

Determine the angle between the initial velocity of the football and the ground.

Possible Answers:

$\theta=33^\circ$ $\displaystyle \theta=33^\circ$

$\theta=77^\circ$ $\displaystyle \theta=77^\circ$

None of these

$\theta=18^\circ$ $\displaystyle \theta=18^\circ$

$\theta=66^\circ$ $\displaystyle \theta=66^\circ$

Correct answer:

$\theta=66^\circ$ $\displaystyle \theta=66^\circ$

Explanation:

Using:

$\displaystyle d_x=v_x*t$

$v_x=\frac{d_x}{t}$ $\displaystyle v_x=\frac{d_x}{t}$

$cos(\theta)=\frac{v_x}{v}$ $\displaystyle cos(\theta)=\frac{v_x}{v}$

Converting mph $\displaystyle mph$ to mps $\displaystyle mps$

$50mph*\frac{.447mps}{1 mph}=22.35\frac{m}{s}$ $\displaystyle 50mph*\frac{.447mps}{1 mph}=22.35\frac{m}{s}$

Converting yards to meters

$50yards*\frac{.914 meters}{1 yard}=45.7 meters$ $\displaystyle 50yards*\frac{.914 meters}{1 yard}=45.7 meters$

Combining equations

$cos(\theta)=\frac{d_x/t}{v}$ $\displaystyle cos(\theta)=\frac{d_x/t}{v}$

Plugging in values:

$cos(\theta)=\frac{45.7/5}{22.35}$ $\displaystyle cos(\theta)=\frac{45.7/5}{22.35}$

$\theta=67^\circ$ $\displaystyle \theta=67^\circ$

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Example Question #55 : Motion In Two Dimensions

$1mph=.447\frac{m}{s}$ $\displaystyle 1mph=.447\frac{m}{s}$

$\displaystyle 1 yd=.914 m$

Determine the initial vector velocity of the football.

Possible Answers:

None of these

$\overrightarrow{v}=<44.1,22.3>\frac{m}{s}$ $\displaystyle \overrightarrow{v}=< 44.1,22.3>\frac{m}{s}$

$\overrightarrow{v}=<9.14,20.4>\frac{m}{s}$ $\displaystyle \overrightarrow{v}=< 9.14,20.4>\frac{m}{s}$

$\overrightarrow{v}=<20.4,9.14>\frac{m}{s}$ $\displaystyle \overrightarrow{v}=< 20.4,9.14>\frac{m}{s}$

$\overrightarrow{v}=<17.5,16.1>\frac{m}{s}$ $\displaystyle \overrightarrow{v}=< 17.5,16.1>\frac{m}{s}$

Correct answer:

$\overrightarrow{v}=<9.14,20.4>\frac{m}{s}$ $\displaystyle \overrightarrow{v}=< 9.14,20.4>\frac{m}{s}$

Explanation:

Determining horizontal component of velocity:

$\displaystyle d_x=v_x*t$

$v_x=\frac{d_x}{t}$ $\displaystyle v_x=\frac{d_x}{t}$

$v_x=\frac{45.7}{5}$ $\displaystyle v_x=\frac{45.7}{5}$

$v_x=9.14\frac{m}{s}$ $\displaystyle v_x=9.14\frac{m}{s}$

Using

$\displaystyle v_x^2+v_y^2=v^2$

Solving for v_y $\displaystyle v_y$

$v_y=\sqrt{v^2-v_x^2}$ $\displaystyle v_y=\sqrt{v^2-v_x^2}$

Combining equations

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$ $\displaystyle v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

Converting mph $\displaystyle mph$ to mps $\displaystyle mps$

$50mph*\frac{.447mps}{1 mph}=22.35\frac{m}{s}$ $\displaystyle 50mph*\frac{.447mps}{1 mph}=22.35\frac{m}{s}$

Converting yards to meters

$50yards*\frac{.914 meters}{1 yard}=45.7 meters$ $\displaystyle 50yards*\frac{.914 meters}{1 yard}=45.7 meters$

Plugging in values:

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$ $\displaystyle v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

$v_y=\sqrt{22.35^2-(\frac{45.7}{5})^2}$ $\displaystyle v_y=\sqrt{22.35^2-(\frac{45.7}{5})^2}$

$v_y=20.4\frac{m}{s}$ $\displaystyle v_y=20.4\frac{m}{s}$

$\overrightarrow{v}=<9.14,20.4>\frac{m}{s}$ $\displaystyle \overrightarrow{v}=< 9.14,20.4>\frac{m}{s}$

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AP Physics 1 : Linear Motion and Momentum

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Example Question #47 : Motion In Two Dimensions

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