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A stone is thrown horizontally at a speed of +5.0m/s off the top of a cliff 78.4m high.
a) how long does it take the stone to reach the bottom of the cliff? (answer 4 sec)
b) how far from the base of the cliff does the stone strike the ground? answer 20 m
c) what are the horizontal and vertical componennts of the velocity of the stone before it hits the ground (vyf=-39.2m/s, vx= 5m/s)
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From my school years, I remember the standard mantra solving this classic Physics problem.
The thrown body participates in two movements.
One movement is horizontal with the horizontal velocity v = 5.0 m/s = constant.
The other movement is vertical uniformly accelerated with the vertical acceleration of g = 9.81 m/s6^2.
The time to get the bottom of the cliff is determined from the equation
= H, or = 78.4 m,
which gives
= = = 15.984; t = = 4 seconds (approximately).
During this time, the horizontal distance is d = v*t = 5.0 * 4 = 20 m, approximately.
Horizontal component of the final velocity is v = 5.0 m (the given constant).
Vertical component of the final velocity is = -g*t = -9.81*4 = -39.2 m/s.
Solved and explained, in all details.