Question

A stone is thrown vertically upward with a velocity of 14m/s.

(a) What is the maximum height will it attain?

(b) What is its velocity on striking the ground?

Answer 1

(a) Let's first find the tme that the stone takes to reach the maximum height:

$$v_y=v_{0y}-gt,$$ $$0=v_{0y}-gt,$$ $$t=\dfrac{v_{0y}}{g}=\dfrac{14\ \dfrac{m}{s}}{9.8\ \dfrac{m}{s^2}}=1.43\ s.$$

Finally, we can find the maximum height attained by the ball:

$$y_{max}=v_{0y}t-\dfrac{1}{2}gt^2,$$ $$y_{max}=14\ \dfrac{m}{s}\cdot1.43\ s-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(1.43\ s)^2=10\ m.$$

b) We can find the velocity of the stone as it strikes the ground from the kinematic equation:

$$v_f^2=v_i^2+2gh_{max},$$ $$v_f=\sqrt{v_i^2+2gh_{max}},$$ $$v_f=\sqrt{0+2\cdot9.8\ \dfrac{m}{s^2}\cdot10\ m}=14\ \dfrac{m}{s}.$$