A ball is projected in a vertical plane with a speed of 20 m/s from th

1.	A ball is projected in a vertical plane with a speed of 20 m/s from the top of a tower of height 84m atan angle of 53° with the horizontal, The time after which it reaches the ground is
Answer» 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:

A ball is projected in a vertical plane with a speed of 20 m/s from the top of a tower of height 84m atan angle of 53° with the horizontal, The time after which it reaches the ground is

Answer»

Answer:INITIAL vertical speed is u=5m/s×sin53

=4m/s

so if the FINAL vertical velocity being v at a height H then v

−2gh will give us v=

16−20×0.45

m/s

the horizontal velocity REMAIN CONSTANT and given as V

=5m/sCos53

=3m/s

so the speed at the said instant will be V=

9+7

=4m/s

Explanation:

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A ball is projected in a vertical plane with a speed of 20 m/s from the top of a tower of height 84m atan angle of 53° with the horizontal, The time after which it reaches the ground is ​

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A ball is projected in a vertical plane with a speed of 20 m/s from the top of a tower of height 84m atan angle of 53° with the horizontal, The time after which it reaches the ground is