Question 1164943
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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.



<pre>
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

    {{{(g*t^2)/2}}} = H,   or  {{{(9.81*t^2)/2}}} = 78.4 m,

which gives

    {{{t^2}}} = {{{(2*H)/g}}} = {{{(2*78.4)/9.81}}} = 15.984;  t = {{{sqrt(15.984)}}} = 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  {{{v[vert]}}} = -g*t = -9.81*4 = -39.2 m/s.
</pre>

Solved and explained, in all details.