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?

Answer 1

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

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

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