Question

A car is parked on a steep incline, making an angle of 37.0° below the horizontal and overlooking the ocean, when its brakes fail and it begins to roll. Starting from rest at t = 0, the car rolls down the incline with a constant acceleration of 4.15 m/s², traveling 52.5 m to the edge of a vertical cliff. The cliff is 30.0 m above the ocean.

(a) Find the speed of the car when it reaches the edge of the cliff.
m/s

(b) Find the time interval elapsed when it arrives there.
s

(c) Find the velocity of the car when it lands in the ocean.

magnitude	m/s
direction	° below the horizontal

(d) Find the total time interval the car is in motion.
s

(e) Find the position of the car when it lands in the ocean, relative to the base of the cliff.
m

Answer 1

a)

Using 3rd equation of motio

V^2= u^2+ 2ad

V^2= 2*4.15*52.5

v= 20.875 m/s

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b)

v= 0+at

t= 20.875/4.15= 5.03 seconds

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c) using 2nd equation of motion

h= u sin 30* t - 0.5gt^2

30= 20.875 sin 37 * t + 4.9t^2

Solving for t

t= 1.5 s

Vertical velocity,

Vy= 20.563 sin 37+ 9.8* 1.5

Vy= 27.08 m/s

Horizontal velocity

Vx= 20.563 cos 37= 16.422 m/s

Net velocity

V^2= Vx^2+ Vy^2

V= 31.67 m/s. (ans)

Direction angle= arctan (Vy/Vx)= 58.77 below horizontal

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d) total time of motion= 1.5+ 5.03= 6.53 seconds

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e)

Horizontal position from cliff

H= 16.422* 1.5= 24.633 m

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Comment in case any doubt.. good luck