# An open lift is moving upward with velocity $10 m/s$. It has an upward acceleration of $2 m/s^2$ . A ball is projected upward with velocity $20 m/s$ relative to ground. Find the time when ball meets the lift

$(a)\;\frac{5}{3}\; seconds \quad (b)\;\frac{5}{2} \;seconds \quad (c)\;\frac{1}{3}\;seconds \quad(d)\;\frac{2}{3} \;seconds$

At a time when ball again meets the lift
Distance covered by lift =Distance covered by ball
$10 t+\large\frac{1}{2}$$\times 2 \times t^2=20 t-\large\frac{1}{2}$$ \times 10 \times t^2$
We get $t=0\; or\; t= \large\frac{5}{3}$
Hence a is the correct answer

edited Jan 26, 2014 by meena.p