A very broad elevator is going up vertically with a constant acceleration of 2 m/s2. 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 30o. The time taken by the ball to return the floor in second is ...Plz explain properly

To solve this problem, we need to analyze the motion of the ball in the context of the accelerating elevator. The elevator is moving upwards with a constant acceleration, and the ball is projected at an angle. We will break down the problem step by step to find the time it takes for the ball to return to the floor of the elevator.

Understanding the Scenario

We have an elevator that is accelerating upwards at 2 m/s². At the moment the ball is projected, the elevator's velocity is 4 m/s. The ball is thrown at an angle of 30 degrees with an initial speed of 4 m/s relative to the elevator's floor.

Breaking Down the Motion

First, we need to determine the components of the ball's initial velocity. Since the ball is projected at an angle, we can use trigonometry to find the vertical and horizontal components of its velocity:

• Vertical component (V_y) = 4 m/s * sin(30°) = 4 m/s * 0.5 = 2 m/s
• Horizontal component (V_x) = 4 m/s * cos(30°) = 4 m/s * (√3/2) ≈ 3.46 m/s

However, since the elevator is accelerating upwards, we must consider the effective acceleration acting on the ball. The ball experiences gravity acting downwards (approximately 9.81 m/s²) and the upward acceleration of the elevator (2 m/s²). Therefore, the effective acceleration acting on the ball is:

Effective Acceleration

The effective acceleration (a_eff) acting on the ball in the upward direction is:

a_eff = g - a_elevator = 9.81 m/s² - 2 m/s² = 7.81 m/s²

Applying the Kinematic Equation

To find the time taken for the ball to return to the floor of the elevator, we can use the kinematic equation for vertical motion:

s = V_y * t - 0.5 * a_eff * t²

Here, s is the displacement, which will be 0 when the ball returns to the floor of the elevator. We can set up the equation as follows:

0 = (2 m/s) * t - 0.5 * (7.81 m/s²) * t²

Rearranging the Equation

Rearranging gives us:

0 = 2t - 3.905t²

Factoring out t:

t(2 - 3.905t) = 0

Finding the Solutions

This gives us two solutions:

• t = 0 (the moment of projection)
• 2 - 3.905t = 0 → t = 2 / 3.905 ≈ 0.512 seconds

Final Result

The time taken by the ball to return to the floor of the elevator is approximately 0.512 seconds. This result shows how the upward acceleration of the elevator affects the motion of the ball, making it take less time to return to the floor than it would in a stationary frame.