Will a Ping-Pong Ball Hit a Doll?

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The discussion revolves around the physics of a ping-pong ball shot from a toy gun aimed at a doll that is nudged off a fence just as the trigger is pulled. Participants explore the equations of motion for both the ball and the doll, considering their respective horizontal and vertical movements. The initial velocity of the ball and the timing of its flight are crucial factors in determining whether it will hit the doll. The outcome depends on how the gun is aimed—whether directly at the doll's initial position or along its projected path. Ultimately, the answer hinges on the precise timing and positioning of both objects during the ball's flight.
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Consider the following: You have a toy gun that shoots ping-pong balls at about 5 m/s. You aim the gun at a doll sitting on a fence. Just as you pull the trigger, a friend of yours (or so you thought) nudged the doll forward, just enough to fall off the fence. Will the ping-pong ball hit the doll? Why or why not?

I'm confused on where to start. So I probably need to assume a distance, correct? Also, do I have to treat the two objects as different projectiles, or just worry about the ping-pong ball?
 
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The distance does not matter, as long as the ground does not ineterfere. It turns out the speed of the ball does not matter either. You can just call the distance d and the speed v_o

Write the equations for the height and the horizontal position of the ball, using initial velocities that correspond to aiming at the doll. Write the equation for the height of the doll. Compare the height of the doll to the height of the ball after the ball has moved a horizontal distance d.
 
so one is in the x direction and the other is in the y direction. ball is in the x and the doll is the y direction.

x = x0 + v0xt
y = y0 +v0yt - 1/2gt^2

or am I looking at this wrong?
 
or actually height would be v0t+(.5)at^2...?
 
courtney1121 said:
so one is in the x direction and the other is in the y direction. ball is in the x and the doll is the y direction.

x = x0 + v0xt
y = y0 +v0yt - 1/2gt^2

or am I looking at this wrong?

The ball moves in both directions. The ball is aimed at the doll, so its initial velocity is resolved into horizontal and vertical components.

courtney1121 said:
or actually height would be v0t+(.5)at^2...?

That's good for the ball. You need the one with y0 for the doll.
 
Last edited:
y = 1/2gt^2?
 
I found the time to be .5s because v/g so the height for the ball is 5*.5 + (.5)*9.8*(.5)^2 which is 6.225m.
 
courtney1121 said:
y = 1/2gt^2?

For the doll, you want the equation you wrote before

y = y0 +v0yt - 1/2gt^2

but recognize that it starts at rest at some initial height h above the starting point of the ball. You will want the same equation for the ball, but starting from a different intial height (perhaps a good place to call zero) with an initial velocity.

courtney1121 said:
I found the time to be .5s because v/g so the height for the ball is 5*.5 + (.5)*9.8*(.5)^2 which is 6.225m.

How could you find a time? There is no specific distance given from the gun to the doll.
 
courtney1121 said:
Consider the following: You have a toy gun that shoots ping-pong balls at about 5 m/s. You aim the gun at a doll sitting on a fence. Just as you pull the trigger, a friend of yours (or so you thought) nudged the doll forward, just enough to fall off the fence. Will the ping-pong ball hit the doll? Why or why not?

Well, the answer depends a bit on what is meant by aiming at the doll. If aiming at the doll means pointing the gun so that the inital velocity of the projectile is directly at the doll, you'll get one answer. If aiming at the doll means pointing the gun so that the current position of the doll is in the projectile's flight path, you might get a different one.
 
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