Question

A tiger leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. how far from the base of the rock will the tiger land?

Answer 1

Let's first find the time that the tiger takes to reach the base of the rock:

$$y=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2y}{g}}=\sqrt{\dfrac{2\cdot7.5\ m}{9.8\ \dfrac{m}{s^2}}}=1.24\ s.$$

Finally, we can find the distance from the base of the rock to the point where the tiger will land:

$$x=v_0t=4.5\ \dfrac{m}{s}\cdot1.24\ s=5.58\ m.$$