A boy throws a ball upwards at $60^{o}$ to the horizontal with a speed of $80 m / s$ and it reaches the maximum height $H$ , If he throws the same ball vertically downward from a height of $H$ with initial speed of $40 m / s$ , The ball will reach ground with speed nearly:

48 m / s

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56 m / s

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64 m / s

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80 m / s

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Solution

The correct option is D $80 m / s$
Maximum height is $H = \frac{v^{2} S i n^{2} θ}{2 g}$
So $H = 240 m$

When the ball is released from thrown from the top then energy will be conserved till this hit the floor so $v_{f} = \sqrt{v^{2} + 2 g h}$

$v_{f} = \sqrt{1600 + 4800} = 80 m / s$

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A boy throws a ball upwards at 60o to the horizontal with a speed of 80 m/s and it reaches the maximum height H , If he throws the same ball vertically downward from a height of H with initial speed of 40 m/s , The ball will reach ground with speed nearly:

The correct option is D 80 m/sMaximum height is H=v2Sin2θ2gSo H=240mWhen the ball is released from thrown from the top then energy will be conserved till this hit the floor so vf=√v2+2ghvf=√1600+4800=80m/s

A boy throws a ball upwards at $60^{o}$ to the horizontal with a speed of $80 m / s$ and it reaches the maximum height $H$ , If he throws the same ball vertically downward from a height of $H$ with initial speed of $40 m / s$ , The ball will reach ground with speed nearly:

The correct option is D $80 m / s$
Maximum height is $H = \frac{v^{2} S i n^{2} θ}{2 g}$
So $H = 240 m$

When the ball is released from thrown from the top then energy will be conserved till this hit the floor so $v_{f} = \sqrt{v^{2} + 2 g h}$

$v_{f} = \sqrt{1600 + 4800} = 80 m / s$