A man throws a rock from the top of a cliff that is 60.0 m above ground level wi

Question

A man throws a rock from the top of a cliff that is 60.0 m above ground level with an initial speed of 35.0 m/s at an angle of 25.0° above the horizontal.

(a) Determine the time it takes for the rock to hit the ground below the cliff.
______ s
(b) Determine the horiztonal distance that the rock travels.
______m
(c) At the instant just before the rock hits the ground below the cliff, find the horizontal and the vertical components of its velocity. (Take up and to the right as positive directions.)

(d) What is the the magnitude of the velocity of the rock just before it hits the ground?
________m/s
(e) What is the angle made by the velocity vector of the rock just before it hits the ground?
________° (below the horizontal)
(f) Find the maximum height above the cliff top reached by the rock.
________ m

Explanation / Answer

Here ,

a) let the time taken is t

Using seocnd equation of motion

y = ut + 0.5 at^2

-60 = 35 * sin(25) * t - 0.5 *9.8 * t^2

solving for t

t = 5.32 s

the time taken for rock to reach ground is 5.32 s

b)

horizontal distance = 35 * cos(25) * 5.32

horizontal distance = 168.8 m

the horizontal distance is 168.8 m

c)

vertical velocity =35 * sin(25) - 9.8 * 5.32

vertical velocity = -37.3 m/s

horizontal velocity = 35 * cos(25)

horizontal velocity =   31.7 m/s

the horizontal velocity is 31.7 m/s

vertical velocity is -37.3 m/s

d)

magnitude of velocity = sqrt(37.3^2 + 31.7^2)

magnitude of velocity = 49 m/s

the magnitude of velocity is 49 m/s just before hitting the ground

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