A cannonball is fired horizontally from the top of a cliff. The cannon is at hei
ID: 2051075 • Letter: A
Question
A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 70.0 m above ground level, and the ball is fired with initial horizontal speed v_0. Assume acceleration due to gravity to be g = 9.80 m/s^2.What is the y position of the cannonball when it is at distance D/2 from the hill? If you need to, you can use the trajectory equation for this projectile, which gives y in terms of x directly:
y = H - rac{g x^2}{2 v_{0 m x}^2}.
You should already know v_0x from the previous part. which is 37 m/s
Explanation / Answer
The equation you need is y=(vo)t + 0.5a(t^2) Though the (vo) here would be the initial velocity in the vertical, y, direction which is zero. So for your case... y = 0.5 a(t^2) You first have to findout how long it takes to hit the ground. Set y = 70 and solve for t. 70 = 0.5 (9.8m/s^2)(t^2) t=3.78 seconds Therefore t/2 = 1.89 seconds Put this back into y = 0.5 a(t^2) and you get y = 17.5 m
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