a soft ball of a mass 0.15kg is dropped, in a vacuum, from a height of 2.0m on to a hard surface. and then the ball bounces
calculate the speed with which the ball hits the ground
Answers
Answered by
8
Answer:
6.32m/s
Explanation:
Velocity of the ball before collision = root 2gH
= root(2×10×2) = 2 root10 ~ 2×3.16 = 6.32 m/s
Answered by
2
Answer:
6.32
Explanation:
mgh = 1/2*m*(v sqaure)
v square = mgh/0.5*m
v square = 0.15x10x2/0.5x0.15 [g = 10, gravitational pull]
v square = 40
v = root{40}
v = 6.32
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