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 speedv_0. Assume acceleration due to gravity to be g = 9.80 m/s^2.

Assume that the cannon is fired at time t = 0 and that the cannonball hits the ground at time t_g. What is the y position of the cannonball at the time tg/2?

I found it tg/2 to be 52.5m

I need help with part b.

Given that the projectile lands at a distance D = 190 m from the cliff, as shown in the figure, find the initial speed of the projectile, v_0.

Answer 1

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