A 200g ball is dropped from a height of 20m , on impact on a ground, it loses 30j of energy. Calculate the height of a ball as reaches on its rebound. (G=10m/s2
energy before hit = m g h
= .2 * 10 * 20
= 40 J
energy after hit = 10 J
m g h = 10
2 h = 10
h = 5 meters
2* 10* 20=40
Mgh=10
2h = 10
h = 5
To find the height of the ball on its rebound, we can use the principle of conservation of energy. The total mechanical energy of a system remains constant as long as there are no external forces acting on it. In this case, we can consider the ball and the Earth as the system.
Let's break down the problem into two parts: the ball falling and the ball rebounding.
1. Ball falling:
The initial potential energy (PE) of the ball can be calculated using the formula PE = mgh, where m is the mass of the ball (200g = 0.2kg), g is the acceleration due to gravity (10m/s^2), and h is the initial height (20m). Therefore, the initial potential energy is:
PE = (0.2kg)(10m/s^2)(20m) = 40J
2. Ball rebounding:
When the ball hits the ground, it loses 30J of energy. This energy loss can be attributed to the conversion of potential energy to other forms (such as kinetic energy, sound, and heat) upon impact. Since the ball rebounds, it still has some potential energy when it reaches its maximum height.
To find the maximum height, we need to determine the final potential energy (PE') of the ball on its rebound. We can use the conservation of energy principle:
Initial potential energy (PE) = Final potential energy (PE') + Energy loss (30J)
Solving for PE':
PE' = PE - Energy loss
PE' = 40J - 30J
PE' = 10J
Finally, we can use the formula PE' = mgh' to find the rebound height (h') of the ball:
10J = (0.2kg)(10m/s^2)h'
h' = 10J / (0.2kg)(10m/s^2)
h' = 5m
Therefore, the height of the ball on its rebound is 5 meters.