Question

A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajectory time from the top of the cliff to the bottom to be 4.3 s. What is the height of the cliff if air resistance is negligibly small?

Answer 1

To solve this problem, it only needs the vertical motion equation for fall free,

$\mathrm{y(t)=\frac{gt^{2}}{2}}$

at    $\mathrm{t=4.3s}$ the distance    $\mathrm{y(t)}$    will be equal to the height $\mathrm{H}$ of the cliff,    $\mathrm{y(4.3s)=H}$ . then

$\mathrm{y(4.3s)=H=\frac{\left( 9.8\tfrac{m}{s^{2}} \right)(4.3s)^{2}}{2}=90.6m}$

the height of the cliff is    $\mathrm{H=90.6m}$