Question:
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 \((k/5) m\). The value of k is ______ (use \(g = 9.8 m/s^2\))
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 \((k/5) m\). The value of k is ______ (use \(g = 9.8 m/s^2\))
Updated On: Oct 21, 2025
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Solution and Explanation
\(v = 19.6 m/s \)
\(t = 6s \)
Time taken in upward motion above tower = \(2s\)
⇒ Time taken from top most point to ground = \(4s\)
⇒ \(\sqrt{\frac{2h}{g}}=4\)
\(h=\frac{16×9.8}{2}=8×9.8\)
⇒ \(k=8×9.8×5=392\)
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