A ball is projected in a vertical plane with a speed of 20 m/s from the top of a tower of height 84m at
an angle of 53° with the horizontal, The time after which it reaches the ground is
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Answer:Initial vertical speed is u=5m/s×sin53
0
=4m/s
so if the final vertical velocity being v at a height h then v
2
=u
2
−2gh will give us v=
16−20×0.45
=
7
m/s
the horizontal velocity remain constant and given as V
x
=5m/sCos53
0
=3m/s
so the speed at the said instant will be V=
V
x
2
+v
2
=
9+7
=4m/s
Explanation:
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