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Grade: 10
        
A body of mass 100 gram is projected vertically up with a velocity of 10 metre per second and another body of same mass is projected vertically down with the same velocity from the top of the tower of height 20 metre calculate the kinetic energy of each ball on reaching the ground take g is equal to 10 metre per second square
5 months ago

Answers : (2)

Arun
22784 Points
							
 
Let the distance covered by the 100g ball before collision be ‘x’.
s = ut+[(1/2)at*t]
s = x, u = 49m/s, a = – 9.8m/(s*s),  t = ?.
Therefore,  x = 49t + [(1/2)*( – 9.8)*t*t]           ...(i)
The height of the tower is 98m 
so, the distance covered by the other ball is 98 – x
Here s = 98 – x,  u = 0, a = 9.8m/(s*s), t = ? (equal to the time of flight of the other ball)
so, 98 – x = 0t + [(1/2)*(9.8)*t*t]
            x = 98 – [(1/2)*(9.8)*t*t]
from (i) we get x = 49t – [(1/2)*(9.8)*t*t]
Therefore 98=49t
and then t = 2s.
and x = 49*2 – [(1/2)*(9.8)*4]
x = 78.4m.

Now after collision

s = 78.4m,  u = 0,  a = 9.8m/(s*s),  t = ?.
Therefore 
78.4 = ½ (9.8)(t*t)
t*t = (78.4)*2 / 9.8
t*t = 16
t = 4.
 
Therefore total time of flight is 6s
5 months ago
Saurabh Koranglekar
askIITians Faculty
3154 Points
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5 months ago
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