Question

A boy standing in an elevator accelerating upwards throws a ball upwards with velocity v the ball return in his hand after a time then the acceleration of the lift is

Answer 1

this must be the answer

Answer 2

Answer:

Acceleration of the lift is $a\ =\ \dfrac{2v\ +\ gt}{t}$

Explanation:

Given,

• Velocity of the ball = v
• Acceleration due to the gravity = g
• Net displacement of the ball = s = 0
• Time taken by the ball to reach in his hand =  t

Now the net displacement covered by the ball during the time interval t is zero duo to the ball where it starts, it ends at that place.

Net acceleration acting on the ball is (g - a) because the lift is accelerating upward,

From the kinematics,

$s\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow 0\ =\ vt\ +\ \dfrac{1}{2}(g\ -\ a}t^2\\\Rightarrow at^2\ =\ 2vt\ +\ gt^2\\\Rightarrow a\ =\ \dfrac{2vt\ +\ gt^2}{t^2}\\\Rightarrow a\ =\ \dfrac{2v\ +\ gt}{t}$

Hence the acceleration of the lift is $\dfrac{2v\ +\ gt}{t}$.