Kinematics

JEE Main level, numerical answer, medium

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Question 162 of 308

A ball is thrown vertically upwards with a velocity of 19.6ms−119.6m{s}^{-1} from the top of a tower. The ball strikes the ground after 6s6s. The height from the ground up to which the ball can rise will be (k5)m\left(\dfrac{k}{5}\right)m. The value of kk is _____ (use g=9.8ms−2g=9.8m{s}^{-2})

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Answer: 392

Initial velocity of ball is u=19.6ms−1u=19.6m{s}^{-1}, final velocity v=0v=0.
Using v=u+atv=u+at, here, acceleration a=−ga=-g.
Time taken in upward motion above tower is ta=ug=19.69.8=2s{t}_{a}=\dfrac{u}{g}=\dfrac{19.6}{9.8}=2s
Now, time taken from top most point to ground is td=6−2=4s{t}_{d}=6-2=4s.
Or, td=2hmax⁡g{t}_{d}=\sqrt{\dfrac{2{h}_{\max }}{g}} (Using s=ut+12gt2s=ut+\dfrac{1}{2}g{t}^{2})
⇒hmax⁡=16×9.82=3925m\Rightarrow {h}_{\max }=\dfrac{16\times 9.8}{2}=\dfrac{392}{5}m.
Hence, the value of k=392k=392.

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