Answer to Question #134336 in Physics for khawla

Question

a motorcyclist launches from a ramp at a horizontal of 20m/s. how far from the base of the cliff does he land if he takes 1.2 seconds to reach the ground?

Answer 1

We can find how far from the base of the cliff does he land from the formula:

$$x = v_0t,$$

here, $x$ is the horizontal displacement of the motorcyclist (or the distance from the base of the cliff to the place where he lands), $v_0$ is the initial velocity of the motorcyclist and $t$ is the time to reach the ground.

Then, from this formula we can calculate how far from the base of the cliff does he land:

$$x = v_0t = 20 \ \dfrac{m}{s} \cdot 1.2 \ s = 24 \ m.$$

Answer:

$x = 24 \ m.$