A ball is thrown vertically upwards with a velocity of 19.6ms^-1 from the top of a tower. The ball strikes the ground after 6s. The height from the…
JEE Main 2022 — Physics Mechanics
2022integermedium
A ball is thrown vertically upwards with a velocity of 19.6ms−1 from the top of a tower. The ball strikes the ground after 6s. The height from the ground up to which the ball can rise will be (5k)m. The value of k is _____ (use g=9.8ms−2)
Official previous-year question
Held on 28 Jul 2022 · Verified 6 Jul 2026.
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Solution
Initial velocity of ball is u=19.6ms−1, final velocity v=0.
Using v=u+at, here, acceleration a=−g.
Time taken in upward motion above tower is ta=gu=9.819.6=2s
Now, time taken from top most point to ground is td=6−2=4s.
Or, td=g2hmax (Using s=ut+21gt2)
⇒hmax=216×9.8=5392m.
Hence, the value of k=392.
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