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Standard XII

Physics

Horizontal and Vertical Components of Projectile Motion

A ball is thr...

Question

A ball is thrown vertically upwards with a velocity of 20 m $s^{- 1}$ from the top of a multistorey building of 25 m high. How high will the ball rise from top of the building? (Take g = 10 m $s^{- 2}$ ).

10 m

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15 m

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20 m

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25 m

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Solution

The correct option is C 20 m
Taking upward direction as $+ v e,$
$u = + 20 m / s a = - g = - 10 m / s^{2}$
At maximum height, $v = 0.$
By third law of motion,
$v^{2} = u^{2} + 2 a s o^{2} = (20)^{2} + 2 (- 10) (s) ⟹ - 400 = 20 s s = 20 m$

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A ball is thrown vertically upwards with a velocity of 20m/s from the top of a multi storey building.The height of the point from where the ball is thrown is 25m from the ground.

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A ball is thrown vertically upwards with a velocity of 20 m s−1 from the top of a multistorey building of 25 m high. How high will the ball rise from top of the building? (Take g = 10 m s−2).

The correct option is C 20 mTaking upward direction as +ve,u=+20m/sa=−g=−10m/s2At maximum height, v=0.By third law of motion,v2=u2+2aso2=(20)2+2(−10)(s)⟹−400=20ss=20m

A ball is thrown vertically upwards with a velocity of 20 m $s^{- 1}$ from the top of a multistorey building of 25 m high. How high will the ball rise from top of the building? (Take g = 10 m $s^{- 2}$ ).

The correct option is C 20 m
Taking upward direction as $+ v e,$
$u = + 20 m / s a = - g = - 10 m / s^{2}$
At maximum height, $v = 0.$
By third law of motion,
$v^{2} = u^{2} + 2 a s o^{2} = (20)^{2} + 2 (- 10) (s) ⟹ - 400 = 20 s s = 20 m$