A ball thrown in vertical upward direction attains maximum height of $16 m$ . At what height would its velocity be half of its initial velocity?

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Solution

Vertical displacement, $h = 16 m$

$v^{2} - u^{2} = 2 a h$

$0 - u^{2} = 2 (- g) h$

$u = \sqrt{2 g h} = \sqrt{2 \times 10 \times 16} = 8 \sqrt{5} m s^{- 1}$

Half of this velocity is $4 \sqrt{5} m s^{- 1}$

$v^{2} - u^{2} = 2 a s$

${(4 \sqrt{5})}^{2} - {(8 \sqrt{5})}^{2} = 2 (- g) H$

$H = 12 m$

Hence, at 12 m velocity of ball is half of initial velocity

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A ball thrown in vertical upward direction attains maximum height of 16m. At what height would its velocity be half of its initial velocity?

Vertical displacement, h=16m v2−u2=2ah 0−u2=2(−g)h u=√2gh=√2×10×16=8√5ms−1 Half of this velocity is 4√5ms−1 v2−u2=2as (4√5)2−(8√5)2=2(−g)H H=12m Hence, at 12 m velocity of ball is half of initial velocity

A ball thrown in vertical upward direction attains maximum height of $16 m$ . At what height would its velocity be half of its initial velocity?