A 0.25 kg skeet (clay target) is fired at an angle of 30 degrees to the horizon with a speed of 25 m/s. When it reaches the maximum height, it is hit from below by a 15 g pellet travelling upward at a speed of 200 m/s. The pellet is embedded in the skeet. (a) How much higher did the skeet go up? (b) How much extra distance does the skeet travel because of the collision?

Question

the image is here:
img85.imageshack.us/img85/905/physicsx.jpg

Answer 1

a.extra height:

Momentum before collision=momentium after collision
but the clay target has no upward momentum at the top, so
.015*200=(.015+.25)V solve for V, that is the new upward velocity.

1/2 mv^2=mgh
h=1/2 v^2/g

now for the extra distance. Figure the time it takes to go h
0=V-9.8t solve for t, then double it (goes back down), that is the extra time in air.
extra distance=2t*25cos30

Answer 2

thanks :)

so a.) 6.5m

in the b.) is the answer 10.07m?

coz i'm not really sure with what i did. but i solved for t using x=v1t+1/2at^2.. then multiplied it with 30cos30. :O