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

Question

A ball of mass 0 . 5 &#160; kg is dropped from the height of 10 &#160; m . The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is _____ m . [Use g = 10 &#160; m &#160; s - 2 ]

Aicharya · Accepted Answer

Consider downward direction to be positive.

Applying first equation of motion, $v = u + a t$

$\Rightarrow v = 0 + g t \Rightarrow g = g t \Rightarrow t = 1 s$

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

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

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

Solution

A ball of mass 0 . 5 &#160; kg is dropped from the height of 10 &#160; m . The height, at which the…

Solution

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