A rock is thrown off a cliff at an angle of 55° above the horizontal. The cliff

Question

A rock is thrown off a cliff at an angle of 55° above the horizontal. The cliff is 135 m high. The initial speed of the rock is 23 m/s. (Assume the height of the thrower is negligible.) (a) How high above the edge of the cliff does the rock rise (in m)? m (b) How far has it moved horizontally when it is at maximum altitude (in m)? m (c) How long after the release does it hit the ground (in s)? s (d) What is the range of the rock (in m)? m (e) What are the horizontal and vertical positions (in m) of the rock relative to the edge of the cliff at t = 2.0 s, t = 4.0 s, and t = 6.0 s? (Assume the +x-direction is in the horizontal direction pointing away from the cliff, the +y-direction is up towards the sky, and x = y = 0 at the point from which the rock is thrown.) x(2.0 s) = m y(2.0 s) = m x(4.0 s) = m y(4.0 s) = m x(6.0 s) = m y(6.0 s) = m

Explanation / Answer

a] Initial vertical velocity = uy = 23sin55 = 18.84 m/s

at the highest point, vy = 0 m/s

so, h = -(0 - 18.842)/(2 x 9.8) = 18.11 m

b] The time it takes to reach this maximum height will be:

v = u + at

=> 0 = 18.84 - 9.8t

=> t = 1.922 s

and since there is no acceleration in the horizontal direction, the distance covered by the rock in the same time will be: Dx = uxt = (23cos55)(1.922) = 25.36 m

c] H = 135 + h = 135 + 18.11 = 153.11 m

use, H = uyt + (1/2)ayt2

153.11 = 0 + (1/2)(9.8)t2

[since at the highest point, the vertical velocity = 0 m/s]

=> t = 5.5898 s

therefore the total time in which the rock comes down will be: T = 1.922 + 5.5898 = 7.512 s.

d] Range of the rock will be:

D = uxT = [23cos55](7.512) = 99.1 m.

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