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Question

# A juggler maintains four balls in the air with air with throwing speed 20 m/s upwards in regular time intervals. When one ball is about to leave his hand the height of balls in air from the ground will be:

A
60m,80m,60m,0m
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B
30m,40m,30m,0m
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C
15m,20m,15m,0m
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D
10m,20m,10m,0m
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Solution

## The correct option is C 15m,20m,15m,0mInitial velocity u=20m/s Final velocity v=0m/s Acceleration a=−10m/s2 Now, the time by ball reach its maximum height From equation of motion v=u+at 0=20−10t t=2s So, the time taken by the same ball to return to the hands of the juggler is =2ug =2×2010 =4s So, he is throwing the balls after 1 sec each, Let at some instant he throws ball number 4. Now, Before 1 s of throwing it, he throws ball 3. So, the height of ball 3 is h3=20×1−12×10×(1)2 h3=15m Before 2s, he throws ball 2, so the height of ball 2 is h2=20×2−12×10×4 h2=20m Before 3s, he throws ball 1, so the height of ball 1 is h1=20×3−12×10×3×3 h3=15m Hence, the height of balls in air from the ground will be 15 m, 20 m, 15 m and 0 m

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