First slide

Relative velocity in 1-d

Question

A very broad elevator plateform is going up vertically with a constant acceleration 1 ms^-2. At the instant when the velocity of the lift is 2 m/s, a stone is projected from the plateform with a speed of 20 m/s relative to the floor at an elevation 30^o. The time taken by the stone to return to the floor will be

Difficult

A

$\frac{30}{11} \sec$
B

$\frac{70}{11} \sec$
C

$\frac{20}{11} \sec$
D

$\frac{90}{11} \sec$

Solution

Initial velocity of stone w.r.t. lift
$\begin{array}{l} = 20 \sin 30^{\circ} \hat{j} + 20 \cos 30^{\circ} \hat{i} \\ = (10 \sqrt{3} \hat{i} + 10 \hat{j}) m / s \end{array}$
Initial velocity of stone w.r.t. ground
$= (10 \sqrt{3} \hat{i} + 12 \hat{j}) m / s$
The initial position of stone and lift are same and when they again meet their final positions will also be same. So both will have same displacement in vertical direction in same time
$Displacement of lift = 2 (t) + \frac{1}{2} \times 1 \times t^{2} = 2 t + \frac{t^{2}}{2}$
$Displacement of stone = 12 (t) - \frac{1}{2} \times 10 \times t^{2} = 12 t - 5 t^{2}$
$so 2 t + \frac{t^{2}}{2} = 12 t - 5 t^{2}$
$\frac{11 t^{2}}{2} = 10 t or t = \frac{20}{11} \sec$
So time taken by stone to return to the floor of lift is $\frac{20}{11} \sec$ .

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