Question

(Figure 1) A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 80.0m above ground level, and the ball is fired with initial horizontal speed v0. Assume acceleration due to gravity to be g = 9.80m/s2 .

Part A

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

Answer numerically in units of meters.

Answer 1

The time it takes a cannonball fired horizontally from a 100m high cliff to reach the ground is just

H = Vyo * t + 1/2 * g * tg^2; (Vyo = 0 for a horizontal shot, solve for tg)

tg^2 = 2 * H / g = 200 / 9.8

tg = 4.5 sec

After half of this time (tg/2 = 2.26 sec), the cannon ball has dropped:

Y = Vyo * t + 1/2 * g * (2.26)^2
Y = 1/2 * 9.8 * (2.26)^2
Y = 25 m (from the top of the cliff)

Notice that the given horizontal initial velocity Vo is irrelevant to the problem.