Question

A ball is thrown horizontally off a cliff 18.2 m/s and hits the ground 30.9 m from the base of the

cliff. How high is the clift?

Answer 1

Answer: 14 m

Step-by-step explanation:

This situation is related to parabolic motion and can be solved by the following equations:

`x=V_(o) cos \theta t` (1)

`y=y_(o)+V_(o) sin \theta t - (g)/(2) t^(2)` (2)

Where:

`x=30.9 m` is the horizontal distance travelled by the ball

`V_(o)=18.2 m/s` is the ball's initial velocity

`\theta=0\°` is the angle (it was thrown horizontally)

`t` is the time

`y=0 m` is the ball's final height

`y_(o)` is the ball's initial height and the cliff height as well

`g=9.8 m/s^(2)` is the acceleration due gravity

Let's begin by finding
`t` from (1):

`t=(x)/(V_(o))` (3)

`t=(30.9 m)/(18.2 m/s)` (4)

`t=1.69 s` (5)

Isolating
`y_(o)` from (2):

`y_(o)=(g)/(2) t^(2)` (6)

Substituting (5) in (6):

`y_(o)=(9.8 m/s^(2))/(2) (1.69 s)^(2)` (7)

Finally:

`y_(o)=13.99 m \approx 14 m`