Explanation:
Given that,
Mass of the ball, m = 1.2 kg
Initial speed of the ball, u = 10 m/s
Height of the floor from ground, h = 32 m
(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :



v = -25.04 m/s (negative as it rebounds)
The impulse acting on the ball is equal to the change in momentum. It can be calculated as :


J = -42.048 kg-m/s
(b) Time of contact, t = 0.02 s
Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :



F = 0.8409 N
Hence, this is the required solution.
Answer:
V = (v1 + v2) / 2 = (8 + 6.5) / 2 = 7.25 m/s average speed
t = 7.2 / 7.25 = .993 sec time to cross patch
a = (v2 - v1) / t = (6.5 - 8) / .993 = -1.51 m/s^2 or 1.5 m/s^2
Answer:
The group that remains unaltered is called the control group.
Answer:
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Explanation: