A stone is thrown horizontally outward from the top of a bridge The stone is released 19.6m above the street below. The initial velocity of the stone is 5.0m/s. Determine the:a)total time that stone is in the air,and b) magnitude of the velocity of the stone just before it strikes the street
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Answer:
If it is thrown horizontal, the horizontal speed u remains 5 m/s until it stops.
u = 5
v = - g t
h = 19.6 - .5 g t^2
or
0 = 19.6 - 4.9 t^2
t^2 = 19.6/4.9
t = 2 seconds in air
u = 5
v = -9.8 * 2 = -19.6 m/s
so speed at ground = sqrt (25 +19.6^2 )
tan angle below horizontal = 19.6/5
Explanation:
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