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A stunt man drives a car at a speed of 20 m/s off a 27-m-high cliff. The road leading to the cliff is inclined upward at an angle of 20 degree.

A) How far from the base of the cliff does the car land?

B) What is the car's impact speed?

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Added Wed, 03 Feb '16
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A) initial velocity will make an angle of 20 degree with horizontal .

in vertical:

h = -27 m

uy = 20sin20 =6.84 m/s

a = -g = -9.8 m/s2

-27 = uy*t + at^2 /2 =6.84 xt - 9.8t^2 /2

4.9t^2 - 6.84t -27 = 0

t =3.14 sec

in horizontal :

distance = horizontal speed x time

= 20cos20 * 3.14 =59.01 m

b) using energy conservation theorem,

mu^2 /2 + mgh = mv^2 /2

20^2 /2 + 9.8x 27 = v^2 /2

v =30.48 m/s

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Added Wed, 03 Feb '16
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