Kinematics

JEE Main level, single correct, medium

160 of 308

Question 160 of 308

A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws nn balls per second, the maximum height the balls can reach is

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Answer: (D) g2n2\dfrac{g}{2{n}^{2}}

As the juggler is throwing nn balls each second and 2nd{2}^{nd} when the first is at its highest point, so the time taken by one ball to reach the highest point, t=1nst=\dfrac{1}{n}s.
And as at highest point v=0v=0.
So, from first equation of motion,
v=u+at⇒0=u−g(1n)⇒u=gnv=u+at \Rightarrow 0=u-g\left(\dfrac{1}{n}\right) \Rightarrow u=\dfrac{g}{n}
Now in order to calculate height, using second equation of motion, we have:
S=ut+12at2\text{S}=ut+\dfrac{1}{2}{\text{at}}^{2} ...(i)

⇒h=1n×gn−12g×(1n)2\Rightarrow h=\dfrac{1}{n}\times \dfrac{g}{n}-\dfrac{1}{2}\text{g}\times {\left(\dfrac{1}{n}\right)}^{2}
⇒h=g2n2\Rightarrow h=\dfrac{g}{2{n}^{2}}
So, the balls rise up to a height of h=g2n2mh=\dfrac{g}{2{n}^{2}}m.

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