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Grade: 11
        
a ball is dropped from a height of 80m on a floor  At each collision with the floor,the ball loses one − tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 8s.
11 months ago

Answers : (2)

Arun
24734 Points
							

Ball is dropped from a height, s = 90 m
Initial velocity of the ball, = 0
Acceleration, = 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)at2
90 = 0 + (1/2) × 9.8 t2

 

t = √18.38  = 4.29 s
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, ur = 9v / 10 = 9 × 42.04 / 10 = 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 × 37.84 / 10 = 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 figure as:

11 months ago
Khimraj
3007 Points
							
it is given that Ball is dropped from a height = 90 m
Time interval, 0 to 12 sec 
Initial velocity of the ball , u =0 m/s
Acceleration due to gravity , g = 9.8 m/s² 
s = ut + 1/2 gt2
where, u = Initial velocity
g = Acceleration due to gravity
s = Distance covered
t = Time
⇒ 90 = 0 + 1/2 × 9.8 × t² 
⇒ 90 = 4.9t² 
⇒ t² = 900/49 
∴ t = 30/7 = 4.28 s 
From 1st equation of motion for freely falling body,
v = u + gt
Where u = 0, g = 9.8 m/s² and t = 4.28s 
⇒ v = 0 + (9.8 ×4.28) ≈ 42 m/s
Bounce velocity of the ball, vb = 0.9v = 37.8 m/s
Time (t’) by the bouncing ball to reach maximum is given by,
v = vb – gt’
where,
v = Final velocity
vb = Bounce velocity
g = Acceleration due to gravity
t’ = Bouncing time
⇒ 0 = 37.84 – (9.8 ×t’)
∴ t’ = 3.86 s
Total time taken by the ball = t + t’ = 4.29 + 3.86 = 8.15 s
 
11 months ago
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