A ball is dropped from a height 2.50 metre above the floors a) find the speed with which it reaches the floor b) the speed of the ball is decreased to 3v/4 due to its collision how high will rise.
Answers
h = 2.5 m
g = 9.8 m/s^2
a. Vt^2 = Vo^2 + 2gh
v^2 = 0 + 2 × 9.8 × 2.5
v^2 = 49
v = √49
v = 7 m/s
b. The new speed v'
= 3v/4
= (3 × 7)/4
= 5.25 m/s
Vt^2 = v'^2 - 2gh
0 = 5.25^2 - 2 . 9.8 . h
0 = 27.5625 - 19.6h
19.6h = 27.5625
h = 27.5625 : 19.6
h = 1.40625 m
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Ball is dropped from a height, s = 90 m
Initial velocity of the ball, u = 0
Acceleration, a = g = 9.8 m/s2
Final velocity of the ball = v
From second equation of motion,
time (t) taken by the ball to hit the ground can be obtained as:
S = ut + 1/2 at^2
90 = 0 + 1/2 x 9.8 x t^2
t = √18.8 = 4.29 sec
From first equation of motion,
final velocity is given as:
v = u + at = 0 + 9.8 × 4.29 = 42.04 m/s
Rebound velocity of the ball,
u(r) =(9/10 )v= (9/10)× 42.04 = 37.84 m/s
Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
v = ur + at′
0 = 37.84 + (– 9.8) t′
t' = -37.84/ -9.8 = 3.86 s
Total time taken by the ball = t + t′ = 4.29 + 3.86 = 8.15 s
As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time. The velocity with which the ball rebounds from the floor
= (9/10)× 37.84= 34.05 m/s
Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s The speed-time graph of the ball is represented in the given attachment ( please see attached file)
I hope, this will help you
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