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QJEEM22Ph7490: JEE Main 2022 Physics - Class 11 JEEM22Ph7490 Solution: A ball is thrown vertically upwards with a velocity of from the …

A ball is thrown vertically upwards with a velocity of 19.6 ms−119.6 \mathrm{~ms}^{-1} from the top of a tower. The ball strikes the ground after 6 s6 \mathrm{~s} . The height from the ground up to which the ball can rise will be (k5)m\left(\frac{k}{5}\right) \mathrm{m} . The value of k\mathrm{k} is __________. (use g=9.8 m/s2\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2} )

Step-by-Step Verified Solution

v = 19.6 m/s

t = 6s

Time taken in upward motion above tower = 2s

⇒\Rightarrow Time taken from top most point to ground = 4s

⇒2hg=4\Rightarrow \sqrt {{{2h} \over g}} = 4

h=16×9.82=8×9.8h = {{16 \times 9.8} \over 2} = 8 \times 9.8

⇒k=8×9.8×5=392\Rightarrow k = 8 \times 9.8 \times 5 = 392

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