question_answer

A very broad elevator is going up vertically with a constant acceleration of\[2\text{ }m/{{s}^{2}}\]. At the instant when its velocity is \[4\text{ }m/s,\]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 to the floor is \[(g=10\text{ }m/{{s}^{2}})\]

A) \[\frac{1}{2}s\]                    

B) \[\frac{1}{3}s\]

C) \[\frac{1}{4}s\]                     

D) \[1\,s\]

Correct Answer: B

Solution :

[b] Components of velocity of ball relative to lift are:             \[{{v}_{x}}=4\cos 30{}^\circ =2\sqrt{3}m/s\]             \[{{v}_{y}}=4sin30{}^\circ =2m/s\] and acceleration of ball relative to lift is \[12\text{ }m/{{s}^{2}}\]in negative y-direction or vertically downwards. Hence time of flight                         \[T=\frac{2{{u}_{y}}}{12}=\frac{{{u}_{y}}}{6}=\frac{2}{6}=\frac{1}{3}s\]