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2 years ago

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.

Physics

1 answer:

yaroslaw [1]2 years ago

5 0

Answer:

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

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