Question

distance travelled by the ball upto that instant, when lift and ball were of same height from ground Take $g=10m/s^2$

• A. $\frac{350}{9}m$
  
• B. $\frac{700}{9}m$
  
• C. $\frac{175}{9}m$
  
• D. None of the above
  

Answer 1

Answer: (C)

$\frac{175}{9}m$

Initial velocity of lift $u_l=10m/s$ (upwards)

Acceleration of the lift $a_l=2m/s^2$ (upwards)

$\therefore$ Height of lift from ground $h=u_{l}t+\frac{a_l}{2}{t}^2$

$\therefore$ $h=10t+\frac{2t^2}{2}$ $⟹h=10t+t^2$ ..........(1)

For ball : Initial velocity of ball $u_b=20m/s$ (upwards)

Acceleration of the ball $a_b=-g=-10m/s^2$ (downwards)

$\therefore$ Height of the ball from ground $h=20t-\frac{10t^2}{2}$

$⟹$ $h=20-5t^2$ .............(2)

From (1) and (2), $10t+t^2=20t-5t^2$ $⟹t=\frac{5}{3}$

$\therefore$ $h=20\times \frac{5}{3}-5\times \frac{5^2}{3^2}=\frac{175}{9}$ m