There is a ball of mass 320 g. It has 625 J potential energy when released freely from a height. The speed with which it will hit the ground is
62·5 m/s
2·0 m/s
50 m/s
40 m/s
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC NDA-2 – 2024
The potential energy at the initial height is PE = 625 J.
Assuming the ball starts from rest, its initial kinetic energy is 0. When it hits the ground, its potential energy becomes 0 (taking the ground as the reference level).
By conservation of energy, Initial Total Energy = Final Total Energy.
PE_initial + KE_initial = PE_final + KE_final
625 J + 0 J = 0 J + KE_final
So, KE_final = 625 J.
The kinetic energy is given by KE = (1/2) * m * v², where v is the speed.
625 = (1/2) * 0.320 * v²
625 = 0.160 * v²
v² = 625 / 0.160 = 625000 / 160 = 62500 / 16
v = sqrt(62500 / 16) = sqrt(62500) / sqrt(16) = 250 / 4 = 125 / 2 = 62.5 m/s.