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Question
A stone is thrown upward with a speed ‘u’ from the top of a tower reaches the ground with velocity ‘3u’. The height of the tower is: (g = acceleration due to gravity)
Options
`(3u^2)/g`
`(4u^2)/g`
`(6u^2)/g`
`(9u^2)/g`
MCQ
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Solution
`bb((4u^2)/g)`
Explanation:
The stone is thrown upwards,
∴ u = −u
Also, v = 3u
Using third kinemetical equation of motion,
v2 = u2 + 2as
As, the stone is in free fall,
a = g and s = h = height of the tower
∴ (3u)2 = (−u)2 + 2gh
∴ 9u2 − u2 = 2gh
∴ 8u2 = 2gh
⇒ h = `(4 u^2)/g`
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