Question

A hoodlum throws a stone vertically downward with an initial speed of 13.0 m/s from the roof of a building, 40.0 m above the ground. (a) How long does it take the stone to reach the ground? (b) What is the speed of the stone at impact?

Answer 1

Answer on Question#51369, Physics, Mechanics | Kinematics | Dynamics

The equations of motion of a stone are $y(t) = H - v_0 t - \frac{g t^2}{2}$ , $v(t) = v_0 + g t$ , where

$$H = 40\,m \quad , \quad v_0 = 13\,\frac{m}{s} \quad , \quad g = 9.81\,\frac{m}{s^2}$$

a) When the stone reaches the ground, $y(t) = 40 - 13t - (9.81)\frac{t^2}{2} = 0$ . Solving this quadratic equation, obtain $t \approx 1.82\,s$ - it takes this time for stone to reach the ground.

b) Using second equation of motion, obtain $v(1.82) \approx 30.85\,\frac{m}{s}$ - that is the speed of the stone at impact.