Question

A BMX rider leaves the end of a horizontal ledge with a velocity of 25 m/s and lands 9.3 m from the base of the ledge. How high is the ledge?

Answer 1

ANSWER

0.68 m

EXPLANATION

The BMX rider leaves the end of the ledge with an initial velocity - which is horizontal, of 25m/s and travels a horizontal distance of 9.3 m until he lands,

In this type of motion, there is no horizontal acceleration so the horizontal velocity is constant, thus the horizontal displacement is,

$x=v_{ox}t$

From this equation, we can find how long the biker was in the air,

$t=\frac{x}{v_{ox}}=\frac{9.3m}{25m/s}=0.372s$

Now we can use this time to find the height of the ledge. The vertical displacement is,

$y=v_{oy}t+\frac{1}{2}gt^2$

The initial vertical velocity is zero because the biker left the ledge horizontally. g is the acceleration due to gravity, 9.81 m/s²,

$y=0+\frac{1}{2}\cdot9.81m/s^2\cdot0.372^2s^2\approx0.68m$

The ledge is 0.68 m high.