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 the ball to reach the floor of the lift is ... (nearly) (g=10 ~ms ⁻² )

Step-by-step solution

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 , aligned 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 aligned