Question

A car traveling 24.5 m/s runs over a cliff and lands 8 m away from the base. how high is the cliff?

Answer 1

Answer: 0.522 m

Step-by-step explanation:

This situation is related to projectile motion, where the initial velocity of the car
`V_(o)=24.5 m/s` has only the horizontal component. This means the angle is equal to zero (
`\theta=0`). And the equations we will use to find the height
`y_(o)` of the cliff are:

`y=y_(o)+V_(o)sen \theta t+(1)/(2)gt^(2)` (1)

Where:

`y=0 m` is the final height of the car

`y_(o)` is the initial height

`V_(o)=24.5 m/s` is the initial velocity of the car

`t` is the time

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

`\theta=0` is the angle

`x=8 m` is the horizontal distance traveled by the car after passing the cliff

Well, firstly we have to find
`t` from (1):

`0=y_(o)+(1)/(2)gt^(2)` (3)

`t=\sqrt{-(2y_(o))/(g)}` (4)

Substituting (4) in (2):

`x=V_(o)cos \theta (\sqrt{-(2y_(o))/(g)})` (5)

Isolating
`y_(o)`:

`y_(o)=-(x^(2) g)/(2V_(o)^(2))` (6)

`y_(o)=-((8m)^(2) (-9.8 m/s^(2)))/(2(24.5)^(2))` (7)

Finally:

`y_(o)=0.522 m`