A ball of a mass m 0.5 kg is dropped from the height hi 1.5 m. It accelerates un

Question

A ball of a mass m 0.5 kg is dropped from the height hi 1.5 m. It accelerates under the influence of gravity with negligible air resistance and rebounds back from the ground to a new height of ha 0.95 m. The time of the impact between the ball and the ground is At 0.0035 s. Find: (a) the impact speed of the ball (in m/s) before it hits the ground (1 point) (b) the take off speed of the ball (in m/s) right after it rebounds back from the ground (1 point): (c) the average force of the impact (in N) (1 point).

Explanation / Answer

given

m = 0.5 kg

hi = 1.5 m

negligible air resistance

rebound height, h2 = 0.95 m

dt = 0.0035 s

a. speed of ball before it hits the ground = u

from work energy theorem

mgh = 0.5mu^2

u = sqrt(2gh) = 29.43 m/s

b. rebound speed = v

from work energy theorem

mgh2 = 0.5mv^2

v = sqrt(2g*hf) = 4.31729 m/s

c. change in momentum = m(v + u) = 16.8736454 kg m/s

hence average force = dP/dt = 4821.041545309 N

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