Question

A motorcycle stunt rider rides off the edge of a cliff. Just

at the edge his velocity is horizontal with magnitude 9.0 m/s. Find the mo-

torcycle’s position, distance from the edge of the cliff, and velocity 0.50 s after

it leaves the edge of the cliff.

Answer 1

This is a question on projectile motion.

The distance from the cliff's edge after motorcycle leaves is the range.

Range = ut

Where u = initial velocity

t = time

u = 9 m/s

t = 0.50 seconds

Range = 0.5 × 9 = 4.5 m

For the vertical components :

We will use the following equation of motion :

V = u + at

Where u = initial velocity

a = acceleration due to gravity

t = time in seconds

V = final velocity

V = 9 + 10 × 0.5 = 9 + 5 = 13

V = 13m/s