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courtney1121
Nov29-06, 10:13 PM
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?

OlderDan
Nov29-06, 10:37 PM
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.

courtney1121
Nov29-06, 10:49 PM
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?

courtney1121
Nov29-06, 10:57 PM
or actually height would be v0t+(.5)at^2...?

OlderDan
Nov29-06, 10:57 PM
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.

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

That's good for the ball. You need the one with y0 for the doll.

courtney1121
Nov29-06, 11:02 PM
y = 1/2gt^2?

courtney1121
Nov29-06, 11:06 PM
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.

OlderDan
Nov30-06, 09:35 AM
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.

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.

NateTG
Nov30-06, 10:21 AM
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.