Quiz 2

Physics 6A

1/22/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

An apple is thrown to a moving person. The apple leaves the thrower's hands $1.5m$ above the ground with a speed of $17.25 m/s$ at an angle $25$ degrees above the horizontal. If the receiver starts $d = 12 m$ away from the thrower along the line of flight of the apple when it is thrown, what constant velocity must the receiver have to get to the apple at the instant it is $1.5m$ above the ground?

Solution

This problem involves two bodies moving at the same time. An apple moving as a projectile with constant acceleration and a person moving at constant velocity. We want to compute the the value of the velocity of the person so that they collide.

We want them to collide at the point where the vertical position of the apple is equal to its initial value.

We can first calculate time it takes for the projectile to again reach $1.5m$ off the ground. Once we have that, we can also calculate its x displacement. Then we can use those two quantities to solve for the velocity of the person so that they collide.

The apple is thrown and caught at the same height off the ground, so the answer won't depend on that $1.5m$. To calculate the time at which it is caught we only need consider the y direction since the x and y direction decouple. Here the initial velocity of the projectile $v_0 =$ 17.25 $m/s$ so if you throw an object up at a velocity $v_{0y} = v_0 \sin(\theta) = 7.29$, then the time to the maximum height, where the the y component of the velocity is zero is given by solving $v = 0 = v_{0y} -gt_m$ where $t_m$ is the time at that maximum. Therefore $t_m = v_{0y}/g = 0.74s$. Therefore by symmetry, the time to return to it's initial height $t_f$ is $t_f = 2t_m = 1.49s$.

Its final $x$ coordinate is $x_f = t_f v_{0x}$ but $v_{0x} = v_0 \cos(\theta) = 15.63$. That is, $x_f = 23.24 m$.

The $x$ coordinate of the person $x_p = d + v_p t$, where $d = 12 m$ and what we want to know. At $t=t_f$, we want that $x_p = x_f$. That yields that equation

$d+v_p t_f = t_f v_{0x}$. Solving we obtain $v_p = v_{0x} - d/t_f = v_{0x} - dg/(2v_{0y}) = 7.56m/s$

Quiz 2

Physics 6A

1/22/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

An apple is thrown to a moving person. The apple leaves the thrower's hands $1.5m$ above the ground with a speed of $15.96 m/s$ at an angle $25$ degrees above the horizontal. If the receiver starts $d = 12 m$ away from the thrower along the line of flight of the apple when it is thrown, what constant velocity must the receiver have to get to the apple at the instant it is $1.5m$ above the ground?

Solution

This problem involves two bodies moving at the same time. An apple moving as a projectile with constant acceleration and a person moving at constant velocity. We want to compute the the value of the velocity of the person so that they collide.

We want them to collide at the point where the vertical position of the apple is equal to its initial value.

We can first calculate time it takes for the projectile to again reach $1.5m$ off the ground. Once we have that, we can also calculate its x displacement. Then we can use those two quantities to solve for the velocity of the person so that they collide.

The apple is thrown and caught at the same height off the ground, so the answer won't depend on that $1.5m$. To calculate the time at which it is caught we only need consider the y direction since the x and y direction decouple. Here the initial velocity of the projectile $v_0 =$ 15.96 $m/s$ so if you throw an object up at a velocity $v_{0y} = v_0 \sin(\theta) = 6.74$, then the time to the maximum height, where the the y component of the velocity is zero is given by solving $v = 0 = v_{0y} -gt_m$ where $t_m$ is the time at that maximum. Therefore $t_m = v_{0y}/g = 0.69s$. Therefore by symmetry, the time to return to it's initial height $t_f$ is $t_f = 2t_m = 1.38s$.

Its final $x$ coordinate is $x_f = t_f v_{0x}$ but $v_{0x} = v_0 \cos(\theta) = 14.46$. That is, $x_f = 19.89 m$.

The $x$ coordinate of the person $x_p = d + v_p t$, where $d = 12 m$ and what we want to know. At $t=t_f$, we want that $x_p = x_f$. That yields that equation

$d+v_p t_f = t_f v_{0x}$. Solving we obtain $v_p = v_{0x} - d/t_f = v_{0x} - dg/(2v_{0y}) = 5.74m/s$

Quiz 2

Physics 6A

1/22/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

An apple is thrown to a moving person. The apple leaves the thrower's hands $1.5m$ above the ground with a speed of $24.66 m/s$ at an angle $25$ degrees above the horizontal. If the receiver starts $d = 12 m$ away from the thrower along the line of flight of the apple when it is thrown, what constant velocity must the receiver have to get to the apple at the instant it is $1.5m$ above the ground?

Solution

This problem involves two bodies moving at the same time. An apple moving as a projectile with constant acceleration and a person moving at constant velocity. We want to compute the the value of the velocity of the person so that they collide.

We want them to collide at the point where the vertical position of the apple is equal to its initial value.

We can first calculate time it takes for the projectile to again reach $1.5m$ off the ground. Once we have that, we can also calculate its x displacement. Then we can use those two quantities to solve for the velocity of the person so that they collide.

The apple is thrown and caught at the same height off the ground, so the answer won't depend on that $1.5m$. To calculate the time at which it is caught we only need consider the y direction since the x and y direction decouple. Here the initial velocity of the projectile $v_0 =$ 24.66 $m/s$ so if you throw an object up at a velocity $v_{0y} = v_0 \sin(\theta) = 10.42$, then the time to the maximum height, where the the y component of the velocity is zero is given by solving $v = 0 = v_{0y} -gt_m$ where $t_m$ is the time at that maximum. Therefore $t_m = v_{0y}/g = 1.06s$. Therefore by symmetry, the time to return to it's initial height $t_f$ is $t_f = 2t_m = 2.12s$.

Its final $x$ coordinate is $x_f = t_f v_{0x}$ but $v_{0x} = v_0 \cos(\theta) = 22.35$. That is, $x_f = 47.49 m$.

The $x$ coordinate of the person $x_p = d + v_p t$, where $d = 12 m$ and what we want to know. At $t=t_f$, we want that $x_p = x_f$. That yields that equation

$d+v_p t_f = t_f v_{0x}$. Solving we obtain $v_p = v_{0x} - d/t_f = v_{0x} - dg/(2v_{0y}) = 16.70m/s$

Quiz 2

Physics 6A

1/22/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

An apple is thrown to a moving person. The apple leaves the thrower's hands $1.5m$ above the ground with a speed of $23.37 m/s$ at an angle $25$ degrees above the horizontal. If the receiver starts $d = 12 m$ away from the thrower along the line of flight of the apple when it is thrown, what constant velocity must the receiver have to get to the apple at the instant it is $1.5m$ above the ground?

Solution

This problem involves two bodies moving at the same time. An apple moving as a projectile with constant acceleration and a person moving at constant velocity. We want to compute the the value of the velocity of the person so that they collide.

We want them to collide at the point where the vertical position of the apple is equal to its initial value.

We can first calculate time it takes for the projectile to again reach $1.5m$ off the ground. Once we have that, we can also calculate its x displacement. Then we can use those two quantities to solve for the velocity of the person so that they collide.

The apple is thrown and caught at the same height off the ground, so the answer won't depend on that $1.5m$. To calculate the time at which it is caught we only need consider the y direction since the x and y direction decouple. Here the initial velocity of the projectile $v_0 =$ 23.37 $m/s$ so if you throw an object up at a velocity $v_{0y} = v_0 \sin(\theta) = 9.88$, then the time to the maximum height, where the the y component of the velocity is zero is given by solving $v = 0 = v_{0y} -gt_m$ where $t_m$ is the time at that maximum. Therefore $t_m = v_{0y}/g = 1.01s$. Therefore by symmetry, the time to return to it's initial height $t_f$ is $t_f = 2t_m = 2.01s$.

Its final $x$ coordinate is $x_f = t_f v_{0x}$ but $v_{0x} = v_0 \cos(\theta) = 21.18$. That is, $x_f = 42.65 m$.

The $x$ coordinate of the person $x_p = d + v_p t$, where $d = 12 m$ and what we want to know. At $t=t_f$, we want that $x_p = x_f$. That yields that equation

$d+v_p t_f = t_f v_{0x}$. Solving we obtain $v_p = v_{0x} - d/t_f = v_{0x} - dg/(2v_{0y}) = 15.22m/s$

Quiz 2

Physics 6A

1/22/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

An apple is thrown to a moving person. The apple leaves the thrower's hands $1.5m$ above the ground with a speed of $22.08 m/s$ at an angle $25$ degrees above the horizontal. If the receiver starts $d = 12 m$ away from the thrower along the line of flight of the apple when it is thrown, what constant velocity must the receiver have to get to the apple at the instant it is $1.5m$ above the ground?

Solution

This problem involves two bodies moving at the same time. An apple moving as a projectile with constant acceleration and a person moving at constant velocity. We want to compute the the value of the velocity of the person so that they collide.

We want them to collide at the point where the vertical position of the apple is equal to its initial value.

We can first calculate time it takes for the projectile to again reach $1.5m$ off the ground. Once we have that, we can also calculate its x displacement. Then we can use those two quantities to solve for the velocity of the person so that they collide.

The apple is thrown and caught at the same height off the ground, so the answer won't depend on that $1.5m$. To calculate the time at which it is caught we only need consider the y direction since the x and y direction decouple. Here the initial velocity of the projectile $v_0 =$ 22.08 $m/s$ so if you throw an object up at a velocity $v_{0y} = v_0 \sin(\theta) = 9.33$, then the time to the maximum height, where the the y component of the velocity is zero is given by solving $v = 0 = v_{0y} -gt_m$ where $t_m$ is the time at that maximum. Therefore $t_m = v_{0y}/g = 0.95s$. Therefore by symmetry, the time to return to it's initial height $t_f$ is $t_f = 2t_m = 1.90s$.

Its final $x$ coordinate is $x_f = t_f v_{0x}$ but $v_{0x} = v_0 \cos(\theta) = 20.01$. That is, $x_f = 38.07 m$.

The $x$ coordinate of the person $x_p = d + v_p t$, where $d = 12 m$ and what we want to know. At $t=t_f$, we want that $x_p = x_f$. That yields that equation

$d+v_p t_f = t_f v_{0x}$. Solving we obtain $v_p = v_{0x} - d/t_f = v_{0x} - dg/(2v_{0y}) = 13.70m/s$

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