Question

A body of mass $4kg$ is thrown upward from a height of $2m$ with an initial speed of $12m/s$, its speed at height $3m$ is ________.

Answer 1

The total mechanical energy of a body = Potential Energy + Kinetic energy
$=mgh+\frac{1}{2}m{v}^2$
On substituting the values of mass, height and velocity of the shot in the formula:
Potential Energy $=4\times 10\times 2$ = $80J$
Kinetic Energy = $0.5\times 4\times {12}^2$ = $288J$
Total energy = Potential energy + Kinetic energy = $368J$
At height = $3m$,
Potential Energy = $4\times 10\times 3=120J$
Kinetic energy = Total energy – Potential energy = $248J$
we know that,
K.E. = $\frac{1}{2}$ m$v^2$
v = $\sqrt{\frac{2\times KE}{m}}$ = $\sqrt{\frac{2\times 248}{4}}$= $11.1m/s$