Question

If a body of mass 3 kg is dropped from the top of a tower of height 250 m then it kinda take energy after 3 second will be

Answer 1

First find the distance covered in 3 sec
$s = \frac{1}{2} \times 10 \times 9 \\ = 45m$
hence
find velocity after 3 sec
${v}^{2} = 2 \times 10 \times 45 \\ {v}^{2} = 900 \\ kinetic \: energy = \frac{1}{2} \times m \times {v}^{2} \\ = \frac{1}{2} \times 3 \times 900 \\ = 1350j$