Question

A ball is thrown horizontally from the roof of a 20.0 m high building and lands 10.0 m away from the base of the building. Determine initial and final velocity.

Answer 1

a) Let's first find the time that the ball takes to reach the ground:

$$y=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2y}{g}}=\sqrt{\dfrac{2\cdot20\ m}{9.8\ \dfrac{m}{s^2}}}=2.0\ s$$

Then, the initial horizontal velocity can be found as follows:

$$v_{0x}=\dfrac{x}{t}=\dfrac{10\ m}{2.0\ s}=5\ \dfrac{m}{s}.$$

b) The initial horizontal velocity remains unchainged during the flight of the ball. Let's find the final vertical velocity of the ball:

$$v_y=v_{0y}-gt=0-9.8\ \dfrac{m}{s^2}\cdot2.0\ s=-19.6\ \dfrac{m}{s}.$$

Then, we can find the final velocity of the ball from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(5\ \dfrac{m}{s})^2+(-19.6\ \dfrac{m}{s})^2}=20.23\ \dfrac{m}{s}.$$