Question

A ball is thrown horizontally at a velocity of 33.0 m/s from a cliff 24.0 meters high. How far from the base of the cliff does the ball land?

Answer 1

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

$$y=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2y}{g}}=\sqrt{\dfrac{2\cdot 24\ m}{9.8\ \dfrac{m}{s^2}}}=2.21\ s.$$

Finally, we can find the distance from the base of the cliff to the place where the ball lands:

$$x=v_0t=33.0\ \dfrac{m}{s}\cdot 2.21\ s=73\ m.$$

Answer:

$x=73\ m.$