A ball at rest is dropped from a height of 12m. It loses 25% of its K.E on striking the ground and bounces back to a height 'h'. then value of 'h' is

Explanation

When the ball is dropped, its initial kinetic energy (KE) is 0 because it's at rest. As it falls, it gains potential energy (PE) due to the increase in height. At the point of impact with the ground, the PE is converted to KE. However, the ball loses 25% of its KE upon striking the ground.Let's denote the initial PE at the top as PE_i and the final PE after bouncing back as PE_f. We know that PE_i = mgh, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of 12m.Here's how to find the height 'h':Calculate the lost KE:Lost KE = 25% of initial KE (since it loses 25%)Lost KE = 0.25 * 0 (as initial KE is 0) = 0 JEnergy Conservation during the fall:PE_i = KE_final (just before hitting the ground)mgh = 0.75 * KE_final (as it loses 25% of its KE)Energy Conservation during the bounce:KE_final (just before hitting the ground) = PE_f (at the highest point after bouncing)0.75 * KE_final = mghCombining the equations:From equation 2 and 3: mgh = 0.75 * mghSince the mass (m) and gravity (g) are constant, we can cancel them out:h = 0.75 * hSolve for h:To find the height 'h' where the ball reaches after bouncing, we need to solve the equation:h = 0.75 * hDividing both sides by 0.75:h / 0.75 = hh = 0.75 * h (since h cannot be 0)Therefore, the ball bounces back to a height of h = 9 meters, which corresponds to answer choice c