Question

A stone of mass $m$ initially at rest and dropped from a height $h$, strikes the surface of the earth with a velocity $v$. If the gravitational force acting on the stone is $W$, then which of the following identities is correct?

• A. $mv-mh=0$
• B. $m{v}^2-W{h}^2=0$
• C. $\frac{1}{2}m{v}^2-Wh=0$
• D. $m{v}^2-mh=0$

Answer 1

Answer: (C)

$\frac{1}{2}m{v}^2-Wh=0$

Conserving total $M.E.$
$mgh+0=0+\frac{1}{2}m{v}^2$

It is given, $mg=W$

So, $\frac{1}{2}m{v}^2-Wh=0$