An object A is kept at the point x=3m and y=1.25m on a plank raised above the ground. At time t=0 the plank starts moving along the +x direction with an accleration 1.5m/s2. At the same instant a stone is projected from the origin with a velocity u as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45 degree to the horizontol. All the motions are in the x-y plane. Find u and the time after which the stone hits the object. Take g=10m/s2.

To solve this problem, we need to analyze the motion of both the stone and the object A on the moving plank. We will break down the problem step by step, focusing on the kinematics involved in both motions.

Understanding the Motion of the Plank and Object A

The object A is located at coordinates (3 m, 1.25 m) on a plank that starts moving in the +x direction with an acceleration of 1.5 m/s². The initial position of the plank is at x = 3 m, and it will continue to move as time progresses.

Position of Object A Over Time

Since the plank accelerates, we can find the position of object A at any time t using the equation of motion:

• x_A(t) = x_0 + v_0 t + 0.5 a t²

Here, x_0 is the initial position (3 m), v_0 is the initial velocity (0 m/s, since it starts from rest), and a is the acceleration (1.5 m/s²). Thus, the position of object A in the x-direction becomes:

• x_A(t) = 3 + 0 + 0.5 * 1.5 * t² = 3 + 0.75t²

Analyzing the Stone's Motion

The stone is projected from the origin (0, 0) with an initial velocity u at an angle of 45 degrees. We can break down this velocity into its horizontal and vertical components:

• u_x = u * cos(45°) = u / √2
• u_y = u * sin(45°) = u / √2

Position of the Stone Over Time

The position of the stone at any time t can be expressed as:

• x_S(t) = u_x * t = (u / √2) * t
• y_S(t) = u_y * t - 0.5 * g * t² = (u / √2) * t - 5t²

Finding the Time of Impact

The stone hits object A when their x and y coordinates are equal. Thus, we set up the following equations:

• x_S(t) = x_A(t)
• y_S(t) = 1.25

From the x-coordinates, we have:

• (u / √2) * t = 3 + 0.75t²

From the y-coordinates, we have:

• (u / √2) * t - 5t² = 1.25

Solving the Equations

First, let's solve the y-coordinate equation for u:

• (u / √2) * t = 1.25 + 5t²

Now, we can express u in terms of t:

• u = √2 * (1.25 + 5t²) / t

Next, substitute this expression for u into the x-coordinate equation:

• (√2 * (1.25 + 5t²) / t√2) * t = 3 + 0.75t²

This simplifies to:

• 1.25 + 5t² = 3 + 0.75t²

Rearranging gives:

• 4.25t² = 1.75

Thus, we find:

• t² = 1.75 / 4.25 = 0.4118
• t ≈ 0.64 seconds

Calculating the Initial Velocity u

Now, substituting t ≈ 0.64 seconds back into the equation for u:

• u = √2 * (1.25 + 5 * (0.64)²) / 0.64

Calculating this gives:

• u ≈ √2 * (1.25 + 5 * 0.4096) / 0.64
• u ≈ √2 * (1.25 + 2.048) / 0.64
• u ≈ √2 * 3.298 / 0.64 ≈ 7.25 m/s

Final Results

To summarize, the initial velocity of the stone is approximately 7.25 m/s, and it will hit the object A after approximately 0.64 seconds.