Question

A rock is thrown from the edge of a cliff to the ground 17.0 m below. The rock has an initial velocity of 20.0 m/s, directed 35.0° above the horizontal. (a) How long does it take the rock to reach the ground?


(b) How far from the base of the cliff does the rock strike the ground?


(c) Find the velocity of the rock just before it strikes the ground.

magnitude
direction	below the horizontal

Answer 1

a)

Considering motion along vertical

h = ho + uy * t - 0.5 gt^2

0 = 17 + (20 sin 35) t - 4.9 t^2

solving for t

t = 3.37 seconds

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b)

Horizontal distance

R = 20 cos 35 * 3.37 = 55.21 m

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c)

vertical velocity

vy = 20 sin 35 - 9.8 * 3.37 = - 21.55 m/s

Horizontal velocity

vx = 2 0 cos 35 = 16.383 m/s

net velocity

v^2 = vx^2 +vy^2

v =27.07 m/s

direction

x = arctan ( vy/vx) = arctan ( 21.55/ 16.383) = 52 below horizontal

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