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Projectile motion equation help

  • Thread starter pkossak
  • Start date
52
0
This question was on my last test, and I got it wrong. If anyone could help me understand how to get the answer, I would really appreciate it!

A small metal ball with a mass of m = 91.7 g is attached to a string of length
l = 1.57 m. It is held at an angle of q = 47.5° with respect to the vertical.
The ball is then released. When the rope is vertical, the ball collides head-
on and perfectly elastically with an identical ball originally at rest. This
second ball flies off with a horizontal initial velocity from a height of h =
3.19 m, and then later it hits the ground. At what distance x will the ball
land?

I'm not even really too sure on how to approach it!
 

Answers and Replies

208
0
you would use projectile motion on this one as well as angular acceleration.

ar = v^2 / r

the range of the projectile isd given by
R = (v062 * sin (2 theta)) / g
 
52
0
Thanks a lot for the help, but is there anything else you can tell me. What's throwing me off is that I'm not sure how to find the velocity. I feel like i'm overlooking something simple :confused:
 
208
0
first you would use the equation

a = g sin (theta)

then intergate to find vel
 
52
0
Got it, thanks so much!
 

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