Question 168 of 308

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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Answer: 5

Consider downward direction to be positive.
Applying first equation of motion, $v=u+at$
$\Rightarrow v=0+gt\Rightarrow g=gt\Rightarrow t=1s$
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 $=\left(10-5\right)m=5m$