A body of mass 'm' is dropped from a height 'h' . The kinetic energy of the body on reaching the ground is ____ .

4

mgh

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

mgh

No worries! We‘ve got your back. Try BYJU‘S free classes today!

mgh

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3

mgh

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C mgh
Initial Energy of the body $=$ K.E+P.E
${K E}_{i} = 0, P E_{i} = m g h$
Final Energy $K E_{f} + P E_{f} P E_{f} = 0$
Energy is conserved
Final Energy $=$ Initial Energy
$K E_{f} = P E_{i}$
$= m g h$

Suggest Corrections

Similar questions

A body dropped from a height H reaches the ground with a speed of 1.2 $\sqrt{g H}$ .Calculate the work doneby air-friction.

1.2 \sqrt{g H}

The potential energy of a body at height h is mgh. Then its kinetic energy just before hitting the ground is

M

h

x m

A body of mass 2 kg is dropped from a building of height 5 m. The maximum kinetic energy the body can attain is ___ J.

Join BYJU'S Learning Program