In a lift moving up with an acceleration of 5 ms-2 a ball is dropped from a height of 1.25 m .The time taken by ball to reach the floor of the lift is........(nearly)(g=10 ms-2)

Question

In a lift moving up with an acceleration of 5 ms-2 a ball is dropped from a height of 1.25 m .The time taken by ball to reach the floor of the lift is........(nearly)(g=10 ms-2) (A) 0.3 second (B) 0.2 second (C) 0.16 second (D) 0.4 second

Tardigrade Admin · Accepted Answer

Distance travelled by ball during fall =1.25 m By using relation s=u t+(1/2) a t2 Here, u=0, s =1.25 m a=(g+a)=(10+5)=15 m / s 2 ⇒ 1.25 =(1/2) × 15 × t2, ⇒ t2=(2 × 1.25/15) t2 =0.16 ⇒ t=0.4 s

Q. In a lift moving up with an acceleration of 5ms−2 a ball is dropped from a height of 1.25m .The time taken by ball to reach the floor of the lift is........(nearly)(g=10ms−2)

0.3 second

0.2 second

0.16 second

0.4 second

Distance travelled by ball during fall =1.25m By using relation s=ut+21​at2 Here, u=0,s=1.25m a=(g+a)=(10+5)=15m/s2 ⇒1.25=21​×15×t2, ⇒t2=152×1.25​ t2=0.16 ⇒t=0.4s

Q. In a lift moving up with an acceleration of $5 m s^{- 2}$ a ball is dropped from a height of $1.25 m$ .The time taken by ball to reach the floor of the lift is........(nearly) $(g = 10 m s^{- 2})$