Question

A stunt car starts at rest at a distance 85m from the edge of a 250m tall cliff. The car accelerates 9 m s toward the / 2 edge of the cliff. As the car leaves the cliff, it falls into the water below. How long did it take to hit the ground?

Answer 1

7.14 s

Step-by-step explanation:

The motion of the car is a projectile motion, which consists of two independent motions:

- A uniform motion along the horizontal direction (constant velocity)

- A uniformly accelerated motion along the vertical direction (free fall)

To find the time of flight of the car, we just analyze the vertical motion, using the suvat equation:

`s=ut+(1)/(2)gt^2`

where:

s = 250 m is the vertical displacement, which is equivalent to the height of the cliff (we take the downward direction as positive direction)

u = 0 is the initial vertical velocity of the car

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

t is the time of flight

And solving for t, we find:

`t=\sqrt{(2s)/(g)}=\sqrt{(2(250))/(9.8)}=7.14 s`