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Time Period of Pendulum

A very broad ...

Question

A very broad elevator is going up vertically with a constant acceleration of $2 m / s^{2}$ . At the instant when its velocity is $4 m / s$ a ball is projected from the floor of the lift with a speed of $4 m / s$ relative to the floor at an elevation of $30^{\circ}$ . The time taken by the ball to return the floor is ( $g$ = $10 m / s^{2}$ ) from time of projection.

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Solution

The correct option is B

$\frac{^{a} b a l l}{l i f t} = a_{b a l l} - a_{l i f t}$
= $- g^j - 2^j$
= $- (g + 2)^j$
= $- 12 \frac{m}{s^{2}}^j$
Now in the lifts frame, the ball is simply a projectile with

a = $- 12 \frac{m}{s^{2}}^j$ and not

a = $= - g \frac{m}{s^{2}}^j$

$∴$ Time of flight = $T$ = $\frac{2 μ s i n θ}{12}$

= $\frac{2 \times 4 \times s i n 30}{12}$

= $2 \times 4 \times \frac{1}{2} \times \frac{1}{12}$

= $\frac{1}{3} s$

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