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

Answer 1

new PE at top=original PE - energy lost
mgh=.2(9.8)20-30
solve for h new.

Answer 2

Solution

Answer 3

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.

Answer 4

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.