Question

If a body of mass 3 kg is dropped from the top of a tower of height 250 m, then its kinetic energy after 3 s will be

• A. $1126J$
• B. $1048J$
• C. $735J$
• D. $1296.5J$

Answer 1

The correct option is D $1296.5J$

Using law of conservation of energy
Gain of kinetic energy after 3s = Decrease in potential energy
Decrease in potential = mgh ....(1)
h is the distance travelled in 3s
From second Equation of motion
$h=\frac{1}{2}g{t}^2....\left(2\right)$

On substituting Equation (2) in Equation (1)

$Decreaseinpotential=\frac{1}{2}m{g}^2{t}^{2}m=3kg,g=9.8m/s^2,t=3sDecreaseinpotential=\frac{1}{2}\times 3\times (9.8{)}^2\times 3^2=1296.5J\text{Thus Kinetic energy after 3s}=1296.5$J