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Question

Two identical balls are shot upward one after another at an interval of 2 s along the same vertical line with same initial velocity of 40 ms1 The height at which the balls collide is

A
50 m
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B
75 m
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C
100 m
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D
125 m
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Solution

The correct option is B 75 m
time taken for ball to rest is
v=u+at
0=40gt
t=4s
therefore ball 1 is at rest at height S=40×412×10×42=80m
velocity of ball 2 after 2s as it was thrown after the delay of 2s
v=u+at
v=4010×2=20m/s
S2=40×212×10×22=60m
now ball 1 will drop with initial velocity of 0 and acceleration g and ball 2 with initial velocity of 20 m/s and deceleration of g and the separation is 20m

S1=0+12gt2
S2=20t12gt2

S1+S2=20
12gt2+20t12gt2=20
t=1s
S1=5m
S2=15m
hence ball 2 will be at 75m when they collide.

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