The fraction of energy that is lost is 25%, it depends how fast the ball was going until it lost 25% of its energy, the gravitational energy was transferred into the kinetic energy that helped the ball bounce back
For more boost and to stop chases of fire
Refer to the diagram shown below.
The basket is represented by a weightless rigid beam of length 0.78 m.
The x-coordinate is measured from the left end of the basket.
The mass at x=0 is 2*0.55 = 1.1 kg.
The weight acting at x = 0 is W₁ = 1.1*9.8 = 10.78 N
The mass near the right end is 1.8 kg.
Its weight is W₂ = 1.8*9.8 = 17.64 N
The fulcrum is in the middle of the basket, therefore its location is
x = 0.78/2 = 0.39 m.
For equilibrium, the sum of moments about the fulcrum is zero.
Therefore
(10.78 N)*(0.39 m) - (17.64 N)*(x-0.39 m) = 0
4.2042 - 17.64x + 6.8796 = 0
-17.64x = -11.0838
x = 0.6283 m
Answer: 0.63 m from the left end.
U have to *modify it to increase its ground clearance*
The time spent in the air by the ball at the given momentum is 6.43 s.
The given parameters;
- <em>momentum of the ball, P = 0.9 kgm/s</em>
- <em>weight of the ball, W = 0.14 N</em>
The impulse experienced by the ball is calculated as follows;

where;
is impulse
is change in momentum
The time of motion of the ball is calculated as follows;

Thus, the time spent in the air by the ball at the given momentum is 6.43 s.
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