To solve the problem of a body thrown perpendicularly from an inclined plane at a 45-degree angle, we need to analyze the motion step by step. The key here is to break down the motion into its components and understand the trajectory of the body as it interacts with the wall and returns to its original position.
Understanding the Motion
When the body is thrown with an initial velocity \( u \) at an angle of 45 degrees to the horizontal, we can decompose this velocity into its horizontal and vertical components. Since the angle is 45 degrees, both components will be equal:
- Horizontal component: \( u_x = u \cos(45^\circ) = \frac{u}{\sqrt{2}} \)
- Vertical component: \( u_y = u \sin(45^\circ) = \frac{u}{\sqrt{2}} \)
Time to Reach the Wall
Next, we need to determine how long it takes for the body to reach the wall. Assuming the wall is at a horizontal distance \( d \) from the point of projection, the time \( t_1 \) to reach the wall can be calculated using the horizontal motion:
Since horizontal velocity is constant, we have:
Distance = Velocity × Time, or \( d = u_x \cdot t_1 \)
Thus, we can express time as:
\( t_1 = \frac{d}{u_x} = \frac{d}{\frac{u}{\sqrt{2}}} = \frac{d \sqrt{2}}{u} \)
Time of Flight Until Impact
During this time \( t_1 \), the body also moves vertically. The vertical motion can be described by the equation:
\( y = u_y t_1 - \frac{1}{2} g t_1^2 \)
Substituting \( u_y \) and \( t_1 \):
\( y = \frac{u}{\sqrt{2}} \cdot \frac{d \sqrt{2}}{u} - \frac{1}{2} g \left( \frac{d \sqrt{2}}{u} \right)^2 \)
After simplifying, we find the vertical displacement just before hitting the wall.
Impact and Return
Upon striking the wall perpendicularly, the body will reverse its horizontal velocity while maintaining its vertical velocity. The time taken to return to the original position after hitting the wall will be equal to the time taken to reach the wall, which is \( t_1 \).
Total Time Calculation
The total time \( T \) for the entire motion, which includes the time to reach the wall and the time to return, can be expressed as:
\( T = t_1 + t_1 = 2t_1 \)
Substituting the expression for \( t_1 \):
\( T = 2 \cdot \frac{d \sqrt{2}}{u} = \frac{2d \sqrt{2}}{u} \)
Final Result
Therefore, the total time period for the body to return to its original position after being thrown from the inclined plane and striking the wall is:
\( T = \frac{2d \sqrt{2}}{u} \)
This formula gives you a clear understanding of how the distance to the wall and the initial velocity affect the total time of flight in this scenario. If you have any further questions or need clarification on any part of this explanation, feel free to ask!
