2.Motion in Straight Line
hard

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

A

$393$

B

$390$

C

$392$

D

$391$

(JEE MAIN-2022)

Solution

$t _{ a }=\frac{ u }{ g }=\frac{19.6}{9.8}=2 s$

$t _{ d }=6-2 s =\sqrt{\frac{2 h _{\max }}{ g }}$

$\Rightarrow h _{\max }=\frac{16 \times 9.8}{2}=\frac{392}{5}$

Standard 11
Physics

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