Question

An elevator is moving up with an acceleration $a{\text{m/s}}^2$. A person inside the lift throws the ball upwards with a velocity $u\text{m/s}$ relative to himself. Then the time of flight of ball ($\text{in sec}$) will be:

• A. $\frac{u}{g+a}$
  
• B. $\frac{u}{g-a}$
  
• C. $\frac{2u}{g+a}$
  
• D. $\frac{2u}{g-a}$
  

Answer 1

The correct option is C $\frac{2u}{g+a}$


The scenario for motion of ball is shown w.r.t lift frame.
with respect to lift frame:
initial velocity of ball $=u_r=u\text{m/s}$
acceleration of ball w.r.t lift: $a_r=-(g+a){\text{m/s}}^2$ , $-ve$ sign implies downward direction.

$\therefore$ applying the equation of motion in lift frame:

$S_r=u_{r}t+\frac{1}{2}{a}_r{t}^2...\left(i\right)$ where $S_r$ is the displacement in the lift frame.

For $t=T=\text{Time of flight}$, putting $S_r=0$ in equation $\left(i\right)$, we get:

$0=uT-\frac{1}{2}(g+a)T^2$

$\therefore T=\frac{2u}{g+a}\text{seconds}$