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 \(g =9.8\,m / s ^{2}\))

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Answer & Solution

Correct Answer

(C) \(392\)

Step-by-step Solution

Detailed explanation

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

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

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

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