A elevator car whose floor to distance is 2.7 m starts ascending with a constant acceleration of 1.2 ~m / s ^2, 2 ~s after a bolt is begin to fall from the ceiling of the car. The free fall time of the bolt is (g=9.8 ~m / s ^e )

Step-by-step solution

Net acceleration on the elevator car is more than acceleration due to gravity. Since elevator car is ascending upwards, from Newton's second law, the net force is acting upwards hence resultant acceleration is a_t=(g+a)=(9.8+1.2)=11 ~m / s ^2 Relative velocity of observer to elevator is u_r=0 from the equation aligned s & =u_r t+ 1 2 a₁ t^2 2.7 & =0+ 1 2 11 t^2 t^2 & = 5.4 11 t & = 5.4 11 ~s aligned