Question

Astone is thrown up reaches a height of 7.2 m before coming down. Find the initial velocity of the stone​

Answer 1

Answer:

12 m/s

Explanation:

Hmax = u^2/2g

7.2 = u^2/20

u^2 = 144

u = 12 m/s

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Answer 2

Answer: 12 m/s

Explanation:

Here, v = 0 (as it started coming down after going up, so at the highest point it should be at rest.)

a = g = -10 m/$s^{2}$

s = h = 7.2 m

By using eq of motion,

$v^{2} = u^{2} + 2as$

$0 = u^{2} + 2as$

$-2as = u^{2}$ [as V=0]

( -2 x -10 x 7.2) = $u^{2}$

∴ $u = \sqrt{144}$

u = 12 m/s

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