a ball of mass 0.2kg is dropped from a height of 20m on impact on the ground it lose 30j of energy.calculate the height which it reaches on the ground and rebound
Data given
Mass = 0.2 kg
Height = 20 m
Energy lost = 30J
Required, Height after rebound = ?
PE after rebound = Original PE - Energy lost
mgh' = mgh - E
0.2×10×h' = 0.2×10×20 - 30
2h' = 40 -30
2h' = 10
h' = 10/2
h' = 5
Therefore, height reached after rebound is 5 m.
new PE at top=original PE - energy lost
mgh=.2(9.8)20-30
solve for h new.
Data given
Mass,m=0.2kg
Height,h=20m
Energy lost,E=30J
Height after rebound=required
from
PE after rebound=Original PE-Energy lost
mgh'=mgh-E
but 'g' is constant, g=10
0.2×10×h'=(0.2×10×20)-30
2h'=40-30
2h'=10
divide by 2 both sides
2h'÷2=10÷2
h'=5
Therefore,the height reached after rebound is 5m.
Why did the ball go to work? Because it needed to bounce back with a new attitude! Let's calculate the rebound height.
First, let's find the initial potential energy of the ball when it was dropped. We can use the formula P.E. = mgh, where:
m = mass of the ball = 0.2 kg
g = acceleration due to gravity = 9.8 m/s²
h = height from which the ball was dropped = 20 m
P.E. = (0.2 kg)(9.8 m/s²)(20 m)
P.E. = 39.2 J
Now, let's subtract the energy lost upon impact from the initial potential energy to find the remaining energy for the rebound. We have:
Remaining energy = Initial potential energy - Energy lost
Remaining energy = 39.2 J - 30 J
Remaining energy = 9.2 J
The remaining energy will be converted into potential energy during the rebound. So, we can equate it to the potential energy formula and solve for the height (h'):
P.E. = mgh'
9.2 J = (0.2 kg)(9.8 m/s²)(h')
h' = 9.2 J / ((0.2 kg)(9.8 m/s²))
h' = 9.2 J / 1.96 kg.m/s²
h' ≈ 4.69 m
So, the ball reaches a height of approximately 4.69 meters on the rebound. Keep bouncing those questions my way!
To calculate the height the ball reaches upon hitting the ground and rebounding, we can use the principle of conservation of energy.
The initial potential energy of the ball when it is at a height of 20m is given by:
Potential energy (PE) = mass (m) * gravitational acceleration (g) * height (h)
PE = 0.2 kg * 9.8 m/s^2 * 20 m
PE = 39.2 J
The ball loses 30 J of energy upon impact and rebound. Therefore, the final potential energy (PE') after rebounding is:
PE' = PE - Energy lost
PE' = 39.2 J - 30 J
PE' = 9.2 J
Now, we can rearrange the formula for potential energy to solve for the height (h') the ball reaches after rebounding:
PE' = m * g * h'
9.2 J = 0.2 kg * 9.8 m/s^2 * h'
Solving for h':
h' = 9.2 J / (0.2 kg * 9.8 m/s^2)
h' ≈ 4.69 m
Therefore, the ball will reach a height of approximately 4.69 meters on the ground after rebounding.