QNEET-2024-Ph-161: NEET 2024 Physics - Class 11 NEET-2024-Ph-161 Solution: An object falls from a height of 10 m above the ground. After…

To solve this problem, we need to consider the principles of energy conservation and the behavior of the object during the rebound. Let's break it down step-by-step: 1. When the object falls from a height of 10   m , it converts its potential energy to kinetic energy at the point of impact. The potential energy (PE) just before hitting the ground can be calculated using the formula: P E = m g h where: m is the mass of the object g is the acceleration due to gravity ( 9.8   m / s 2 ) h is the height ( 10   m ) The kinetic energy (KE) of the object just before impact is equal to the potential energy it had at the height of 10   m , because of the conservation of energy principle. 2. Upon hitting the ground, the object loses 50 % of its kinetic energy. Therefore, the kinetic energy after the impact is: K E a f t e r   i m p a c t = 1 2 K E b e f o r e   i m p a c t 3. The object will now rebound to a height where its kinetic energy is converted back to potential energy. Let this height be h r e b o u n d . The potential energy at this rebound height is: P E r e b o u n d = m g h r e b o u n d Since the kinetic energy after the impact is 1 2 m g h , we set this equal to the potential energy at the rebound height: 1 2 m g h = m g h r e b o u n d By canceling out the mass m and the gravitational constant g , we get: 1 2 h = h r e b o u n d Substituting h = 10   m , we find: 1 2 _ 10   m = h r e b o u n d h r e b o u n d = 5   m Therefore, the height up to which the object can rebound from the ground is 5   m . The correct answer is: Option D 5   m