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Consider a seesaw of length $l = 8 m$ and height $h = 2 m$ as shown. A small boy sits at one end of the seesaw. Another body pushes down the other end of the seesaw. At the instant when the other end hits the ground, the angular velocity of the seesaw is $\sqrt{g}$ and the small boy is airborne. The maximum height reached by the small boy is $\frac{5 a}{3}$ meters. The value of a is__

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answer is 6.

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Detailed Solution

$u = ω r = \frac{ω l}{2} = 4 \sqrt{g}$

$\cos θ = \frac{2 \sqrt{3}}{4} = \frac{\sqrt{3}}{2}$

$α = \frac{π}{2} - θ$

maximum height from the level of B:

$H = \frac{u^{2} \sin^{2} α}{2 g} = \frac{4^{2} g \cos^{2} θ}{2 g} = 6 m$

maximum height from level of ground (AC):

$H_{m a x} = 2 h + H = 10 m$

$\Rightarrow \frac{5 a}{3} = 10 \Rightarrow a = 6$

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