Question

A lift whose floor to ceiling distane is 2.50m starts ascending with a constant accleration of 1.25m/s . Obe secind after the start , a bolt begins to fall frim the ceiling of the lift . The time which the bolt hits the floor is (take g =10m/s).

Answer 1

Answer on Question #42976, Physics, Mechanics | Kinematics | Dynamics

A lift whose floor to ceiling distance is 2.50m starts ascending with a constant acceleration of $1.25\mathrm{m / s}$ . One second after the start, a bolt begins to fall from the ceiling of the lift. The time which the bolt hits the floor is (take $\mathsf{g} = 10\mathsf{m / s}$ ).

Solution:

Let us consider the line of motion of elevator and bolt as the Y-axis and the floor's initial position (when the bolt starts falling) as origin.

At the moment when the bolt starts falling, speed of the elevator and the bolt is

$$v _ {1} = v _ {0} + a t = 0 + 1. 2 5 \cdot 1 = 1. 2 5 \mathrm {m / s}$$

Let $t_1$ be the time after which the bolt strikes the floor.

The y-coordinate of the bolt at time $t_1$ is

$$y _ {b o l t} = y _ {0} + v _ {1} t _ {1} + \frac {a t _ {1} ^ {2}}{2} = y _ {0} + v _ {1} t _ {1} - \frac {g t _ {1} ^ {2}}{2} = 2. 5 + 1. 2 5 t _ {1} - \frac {1 0 t _ {1} ^ {2}}{2}$$

(As the bolt is freely falling, its acceleration is -g).

The y-coordinate of the floor at time $t_1$ is

$$y _ {f l o o r} = y _ {0} + v _ {1} t _ {1} + \frac {a t _ {1} ^ {2}}{2} = 0 + 1. 2 5 t _ {1} + \frac {1 . 2 5 t _ {1} ^ {2}}{2}$$

As the bolt strikes the floor at time $t_1$ , $y_{\text{bolt}} = y_{\text{floor}}$

Thus,

$$2. 5 + 1. 2 5 t _ {1} - \frac {1 0 t _ {1} ^ {2}}{2} = 0 + 1. 2 5 t _ {1} + \frac {1 . 2 5 t _ {1} ^ {2}}{2}$$ $$t _ {1} ^ {2} = \frac {5}{1 0 + 1 . 2 5} = \frac {5}{1 1 . 2 5} = 0. 4 4 4$$ $$t _ {1} = \sqrt {0 . 4 4 4} = 0. 6 7 \mathrm {s}$$

Answer: $t_1 = 0.67$ s.

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