A very broad elevator is going up vertically | vedclass

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(B) To find the time taken by the ball to return to the floor,we analyze the motion relative to the elevator.Let the upward direction be positive.The

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$1/3$

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(B) To find the time taken by the ball to return to the floor,we analyze the motion relative to the elevator.Let the upward direction be positive.The acceleration of the elevator is $a_e = +2\,m/s^2$.The acceleration of the ball relative to the ground is $a_b = -g = -10\,m/s^2$.The relative acceleration of the ball with respect to the elevator is $a_r = a_b - a_e = -10 - 2 = -12\,m/s^2$.The vertical component of the initial velocity of the ball relative to the floor is $u_y = u \sin 	heta = 4 \sin 30^{\circ} = 4 	imes 0.5 = 2\,m/s$.Using the equation of motion for displacement $s = u_y t + \frac{1}{2} a_r t^2$,for the ball to return to the floor,the displacement $s$ relative to the elevator must be $0$.$0 = 2t + \frac{1}{2}(-12)t^2$$0 = 2t - 6t^2$$6t^2 = 2t$Since $t 
eq 0$,we have $t = \frac{2}{6} = \frac{1}{3}\,s$.

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