Question

A truck driver named gabriel went crazy and decided to ride off the edge of a cliff. Just at the edge, his velocity is horizontal with magnitude of 12.0m/s. Find his position, distance fromthe edge of the cliff. Find the velocity after .50 seconds.

Answer 1

t = 0.50 s

$v_{0}=12 m/s$

This is a question on projectile motion. 

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

$range=v_{0}t$

Where $v_{0}$ = initial velocity

t = time

Range = 0.5 × 12 = 6 m

For the vertical components :

We will use the following equation of motion :

$v=v_{0}+at$

a = acceleration due to gravity

$v$ = final velocity

$v=12+5=17 m/s$