A ball of mass 0 . 5kg is dropped from the height of 10m . The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is _____ m . [Use g = 10m s ^ - 2 ]

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A ball of mass 0 . 5kg is dropped from the height of 10m . The height, at which the magnitude of velocity becomes equal

Consider downward direction to be positive.

Applying first equation of motion, $v = u + at$

$⇒ v = 0 + gt ⇒ g = gt ⇒ t = 1s$

Distance covered in $1s$ , $s = ut + \dfrac{1}{2}a{t}^{2} = 0 + \dfrac{1}{2}g \times {1}^{2} = \dfrac{10}{2} = 5m$

Hence, height of the point from the surface $= \left(10 - 5\right)m = 5m$

A ball of mass 0 . 5kg is dropped from the height of 10m . The...

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