Answer to Question #162817 in Mechanics | Relativity for M

Question

A ball thrown up from a rooftop of height 54m lands on the ground in 5s. What is its speed 13m below?

Answer 1

Let's first find the initial speed of the ball from the kinematic equation:

$$y=y_0+v_0t-\dfrac{1}{2}gt^2,$$ $$0=54+5v_0-122.5,$$ $$v_0=13.7\ \dfrac{m}{s}.$$

Let's find the time that the ball takes to reach the height 13 meters below rooftop:

$$41=54+13.7t-4.9t^2,$$ $$4.9t^2-13.7t-13=0.$$

This quadratic equation has two roots: $t_1=3.54\ s$ and $t_2=-0.75\ s$. Since time can't be negative, the correct answer is $t=3.54\ s$.

Finally, we can find the speed of the ball 13 meters below rooftop:

$$v=v_0-gt,$$ $$v=13.7\ \dfrac{m}{s}-9.8\ \dfrac{m}{s^2}\cdot3.54\ s=-21\ \dfrac{m}{s}.$$

The sign minus means that the speed of the ball directed downward.