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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,0m

Initial 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=2010t

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=20×112×10×(1)2

h3=15m

Before 2s, he throws ball 2, so the height of ball 2 is

h2=20×212×10×4

h2=20m

Before 3s, he throws ball 1, so the height of ball 1 is

h1=20×312×10×3×3

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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