Question

A stunt man drives a car at a speed of 20 m/sm/s off a 32-mm-high cliff. The road leading to the cliff is inclined upward at an angle of 20∘∘.

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

B, What is the car's impact speed?

Answer 1

initial velocity will make an angle of 20 degree with horizontal .

in vertical:

h = -27 m

u_y = 20sin20 =6.84 m/s

a = -g = -9.8 m/s2

-32 = u_yt + at²/2 =6.84t - 9.8t² /2

4.9t² - 6.84t -32 = 0

t =3.35 sec

in horizontal :

distance = horizontal speed x time

= 20cos20 x 3.35 =62.96 m (range)

b) using energy conservation theorem,

mu² /2 + mgh = mv² /2

20^² /2 + 9.8x 32 = v² /2

v =32.05 m/s (impact velocity)

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