Question

a ball was kicked horizontally off a clif at 15m/s, how high was the cliff if the ball landed 83 m from base of the cliff

Answer 1

The ball's horizontal position $x$ at time $t$ is

$x=v_{0x}t$

The ball has initial velocity in the horizontal direction only, so $v_{0x}=15\,\frac{\mathrm m}{\mathrm s}$. Then the time it takes for the ball to travel 83 m horizontally $t$ is given by

$83\,\mathrm m=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)t\implies t=5.5\,\mathrm s$

Meanwhile, the ball's vertical position $y$ at time $t$, starting at the height of the cliff $h$, is given by

$y=h-\dfrac12gt^2$

where $g=9.8\,\frac{\mathrm m}{\mathrm s^2}$ is the acceleration due to gravity. After 5.5 seconds, the ball hits the ground, so that $y=0$ when $t=5.5\,\mathrm s$, and we use this to solve for $h$:

$0=h-\dfrac12g(5.5\,\mathrm s)^2\implies h\approx150\,\mathrm m$