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\,(\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\,(\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-\frac12gt^2`

where
`g=9.8\,(\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-\frac12g(5.5\,\mathrm s)^2\implies h\approx150\,\mathrm m`