Question

A cannon ball fired horizontally from a cliff has a velocity directed at 60 degrees above horizontal when it hits the ground 3.0 seconds later. how high is the cliff

Answer 1

Answer: 44.1 m

Explanation:

The canon ball is fired horizontally. Hence, the initial velocity in vertical direction, u=0.

It takes 3.0 s for the ball to hit the ground. t=3.0 s

the ball would experience acceleration due to gravity i.e. a=g=9.8 m/s²

The vertical distance covered = height of the cliff.

Using the equation of motion,

$s=ut+\frac{1}{2}at^2$

$\Rightarrow h=0+\frac{1}{2}9.8m/s^2\times (3.0s)^2=44.1 m$

Hence, the height of the cliff is 44.1 m.

Answer 2

Since the ball is fired horizontally, the initial y velocity is zero and the time to hit the ground is the same as if the ball was simply dropped from the cliff.  So you can solve the y position function:
$-\frac{1}{2}g t^{2} + h_{o}=0$
giving a height of 44.1m.
The given final velocity vector tells us that the initial x-directed velocity was about 17m/s.