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Second Law of Motion

A lift is mov...

Question

A lift is moving up with acceleration a. A person inside the lift throws the ball upwards with a velocity u relative to hand.
a.What is the time of flight of the ball?
b.What is the maximum height reached by the ball in the lift?

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Solution

$a^{''}$ (acceleration with respect to lift) $= a + g$

Since ball comes back, $y = 0 \to y = \to μ t + \frac{1}{2} \to g t^{2} \Rightarrow 0 = μ t + \frac{1}{2} - (g + a) t^{2} \Rightarrow \frac{2 μ}{(g + a)} = t$

Height reached by ball $0 - μ^{2} = 2 (- (g + a)) H \Rightarrow H = \frac{μ^{2}}{2 (g + a)}$

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