A person standing at the edge of a seaside cliff kicks a stone over the edge wit

Question

A person standing at the edge of a seaside cliff kicks a stone over the edge with a velocity of vi = 25 m/s. The cliff is h = 46 m above the water's surface

(a) How long does it take for the stone to fall to the water?
s

(b) What is its horizontal velocity when it strikes the water?
m/s

(c) What is its vertical velocity when it strikes the water?
m/s

(d) With what impact velocity does it strike the water?
Magnitude m/s

(e) At what angle (clockwise below +X expressed as a positive number) is it moving when it strikes the water?
°

Explanation / Answer

a. T= (2h/g)1/2

taking g=10m/s2

=(2 x 46/10)

3.033secs

b. horizontal velocity is same throughout and hence it is 25m/s

c. vertical velocity = gt= 10 x 3.033s=30.33m/s

d. it strikes the ground with [u2 + (gt)2]1/2

=26.77m/s

e. angle at which it is moving is  = tan-1 (gt/u)

=tan-1 (10 X 3.033/25)

=50.50 degrees

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