A small cannonball is shot straight up from the ground with a launch speed of 26
ID: 1583622 • Letter: A
Question
A small cannonball is shot straight up from the ground with a launch speed of 26.3 m/s. 2.00 seconds after the cannonball is launched, a ping pong ball is dropped from a height of 36.5 m. Both objects only move in the y-direction.
2a. Where (with respect to the ground) do the cannonball and the ping pong ball pass one another?
2b. How fast and in what direction are each of the balls traveling when they pass one another?
2c. Which ball hits the ground first? How long after the first ball hits the ground does the second ball hit the ground
Explanation / Answer
2A. suppose they pass t time after ping pong ball released,
y_ping pong = 36.5 - g t^2 /2
= 36.5 - 4.9t^2
for cannonball,
y = 26.3(t + 2) - 9.8(t + 2)^2 / 2
height must be same,
36.5 - 4.9t^2 = 26.3t + 52.6 - 4.9 t^2 -19.6t - 19.6
36.5- 52.6 + 19.6 = 26.3t - 19.6t
t = 0.522 sec
from ground, h = 36.5 - 4.9t^2 = 35.2 m ......Ans
2b) pingpong ball,
v = - 9.8 x 0.522 = - 5.12 m/s
So 5.12 m/s downward
cannonball, v = 26.3 - (9.8)(2 +0.522)
= 1.58 m/s
So 1.58 m/s upward
2c) after that ,
for pingpong ball,
0 - 35.2 = - 5.12t - 4.9t^2
t = 2.21 sec
fro cannonball,
0 - 35.2 = 1.58 t - 4.9 t^2
t = 2.85 sec
so pingpong ball will reach first.
time difference = 2.85 - 2.21 = 0.64 sec