A ball of mass 0.5 kg 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/s²]

Let ‘h’ be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

$h=ut+\frac{1}{2}a{t}^2$

$=0\times 1+\frac{1}{2}\times 10\times 1\times 1$

= 5m

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