Question

A stunt driver drives a car at 15 m/s off a cliff and into a lake. The surface of the water is 25 m below the cliff.

a) How long did it take to reach the water?
b) What was the horizontal distance from the cliff where the car entered the water?
c) At what velocity did the car hit the water?

Answer 1

Step-by-step explanation:

Given

initial velocity u = 15m/s

Height below the water = 25m

a) Using the equation of motion S = (v+u)/2 * t

25 = 0+15/2 * t

25 = 7.5t

t = 25/7.5

t = 3.33s

Hence it will take 3.33secs to reach the water

b) The horizontal distance is expressed as;

S = Uxt

S = 15(3.33)

S = 50m

Hence the horizontal distance from the cliff where the car entered the water is 50m

c) The velocity of the car v.

Using the equation of motion;

v² = u²+2gS

v² = 15²+2(9.8)(25)

v² = 225+490

v = √715

v = 26.74m/s

Hence the car hit the water at the velocity of 26.74m.s