Answer to Question #262259 in Physics for Jey Jeremy

Question

A stone is thrown straight downward with initial speed 8.0 m/s from a height of 25 m. Find (a) the time it takes to reach the ground and (b) the speed with which it strikes.

Answer 1

(a) Let's take the downwards as the positive direction. We can find the time that the stone takes to reach the ground from the kinematic equation:

$$d=v_0t+\dfrac{1}{2}gt^2,$$ $$25=8t+4.9t^2,$$ $$4.9t^2+8t-25=0.$$

This quadratic equation has two roots: $t_1=1.59\ s$ and $t_2=-3.22\ s.$ Since time can't be negative the correct answer is $t=1.59\ s$.

(b) We can find the speed with which the stone strikes the ground from the kinematic equation:

$$v=v_0+gt=8\ \dfrac{m}{s}+9.8\ \dfrac{m}{s^2}\times1.59\ s=23.6\ \dfrac{m}{s}.$$