A tiger leaps horizontally from a 12 m high rock with a speed of 4.5 m/s. How far from the base of the rock will they land?

Answer 1

In a free-fall question such as this, the horizontal velocity does not change and the horizontal distance is velocity x time. The general equation of motion is x = x0 + v0 t + (1/2) a t²

In the x direction this reduces to x = v0 t, setting the start as x=0 and because a_x = 0

In the y direction this reduces to y = 12 - (1/2) (9.8m/s²) t² because the rock is 12 m high, ground level = 0, and a = g. You could also set this up so that down is positive and the starting point is y=0. Won't change the answer.

Solve the y-equation for t ( t = √(2y/g) ), then use the horizontal velocity given and this calculated time to find the distance.