Question

A rock is dropped from a 100-m-high cliff. How long does it take to fall (a) the first 50 m and (b) the second 50 m?

Answer 1

Taking the $+y$ direction downward and $y_0$ = 0, we have $y$ = $V_{0}t+\frac{1}{2}g{t}^2$, which (with $V_0$ = 0) yields $t$ = $\sqrt{2}y/g$(a) For this part of the motion, $y_1$ = 50 m so that $t_1$ = $\sqrt{\frac{2\left(50m\right)}{9.8m/s^2}}$ = 3.2 s. (b) For this next part of the motion, we note that the total displacement is $y_2$ = 100 m. Therefore, the total time is $t_2$ = $\sqrt{\frac{2\left(100m\right)}{9.8m/s^2}}$ = 4.5 s. The difference between this and the answer to part (a) is the time required to fall through that second 50 m distance: $\Delta t$ = $t_2-t_1$ = 4.5 s – 3.2 s = 1.3 s.