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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 msâˆ’1 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 mtime taken for ball to rest is v=u+at0=40−gtt=4stherefore ball 1 is at rest at height S=40×4−12×10×42=80m velocity of ball 2 after 2s as it was thrown after the delay of 2sv=u+atv=40−10×2=20m/sS2=40×2−12×10×22=60mnow 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 20mS1=0+12gt2S2=20t−12gt2S1+S2=2012gt2+20t−12gt2=20t=1sS1=5mS2=15m hence ball 2 will be at 75m when they collide.

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