Asked by Thobiswa
A ball of mass 200 g is dropped from a height 3 m above the ground onto a hard floor, and rebounds to a height of 2.2 m. If the ball is in contact with the floor for 0.002 s calculate the average force exerted by the floor on the ball
Answers
Answered by
Henry
V^2 = Vo^2 + 2g*h = 0 + 19.6*3 = 58.8
V = 7.67 m/s.
V^2 = Vo^2 + 2g*h = 0 @ max ht.
Vo^2 - 19.6*2.2 = 0
Vo^2 = 43.12
Vo = 6.57 m/s.
a = (-6.57-7.67)/0.002s = -7120 m/s^2.
F = m*a = 0.20 * (-7120) = -1424 N.
V = 7.67 m/s.
V^2 = Vo^2 + 2g*h = 0 @ max ht.
Vo^2 - 19.6*2.2 = 0
Vo^2 = 43.12
Vo = 6.57 m/s.
a = (-6.57-7.67)/0.002s = -7120 m/s^2.
F = m*a = 0.20 * (-7120) = -1424 N.
Answered by
UEW
A ball of mass 1kg is dropped from a height of 7m and rebounds to a height of 4.5m. Calculate:
(I) its KE just before impact
(II) its initial rebound velocity and KE
(III) Account for the loss of KE on the impact.
(I) its KE just before impact
(II) its initial rebound velocity and KE
(III) Account for the loss of KE on the impact.
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