To tackle this problem, we need to analyze the motion of both bodies: the first one that falls freely from a height \( H \) and the second one that is thrown from the ground. The goal is to find the initial velocity and angle of projection of the second body so that it collides with the first body at an altitude of \( h = \frac{H}{2} \) while maintaining a horizontal distance \( l \) between them. Let's break this down step by step.
Understanding the Motion of the Falling Body
The first body is in free fall, which means it is only influenced by gravity. The equations of motion for the falling body can be expressed as:
- Vertical displacement: \( y = H - \frac{1}{2} g t^2 \)
- Where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
We want to find the time \( t_f \) it takes for the first body to fall to the height \( h = \frac{H}{2} \). Setting \( y = \frac{H}{2} \), we have:
\( \frac{H}{2} = H - \frac{1}{2} g t_f^2 \)
Rearranging gives:
\( \frac{1}{2} g t_f^2 = H - \frac{H}{2} = \frac{H}{2} \)
Thus, we find:
\( g t_f^2 = H \) or \( t_f^2 = \frac{H}{g} \)
From this, we can derive:
\( t_f = \sqrt{\frac{H}{g}} \)
Analyzing the Thrown Body
The second body is thrown with an initial velocity \( v_0 \) at an angle \( \theta \) from the horizontal. The vertical and horizontal components of its motion can be described as follows:
- Vertical motion: \( y = v_0 \sin(\theta) t - \frac{1}{2} g t^2 \)
- Horizontal motion: \( x = v_0 \cos(\theta) t \)
Since we want the second body to collide with the first at \( h = \frac{H}{2} \) when the first body has fallen for time \( t_f \), we set \( t = t_f \) in the equations for the second body:
For vertical motion:
\( \frac{H}{2} = v_0 \sin(\theta) t_f - \frac{1}{2} g t_f^2 \)
Substituting \( t_f \) gives:
\( \frac{H}{2} = v_0 \sin(\theta) \sqrt{\frac{H}{g}} - \frac{1}{2} g \left(\frac{H}{g}\right) \)
Which simplifies to:
\( \frac{H}{2} = v_0 \sin(\theta) \sqrt{\frac{H}{g}} - \frac{H}{2} \)
Rearranging yields:
\( v_0 \sin(\theta) \sqrt{\frac{H}{g}} = H \)
Thus, we can express \( v_0 \sin(\theta) \) as:
\( v_0 \sin(\theta) = \sqrt{\frac{gH}{4}} \)
Horizontal Motion Considerations
For the horizontal motion, we have:
Since the horizontal distance traveled is \( l \), we can write:
\( l = v_0 \cos(\theta) t_f \)
Substituting \( t_f \) gives:
\( l = v_0 \cos(\theta) \sqrt{\frac{H}{g}} \)
Solving for Initial Velocity and Angle
Now we have two equations:
- From vertical motion: \( v_0 \sin(\theta) = \sqrt{\frac{gH}{4}} \)
- From horizontal motion: \( v_0 \cos(\theta) = \frac{l \sqrt{g}}{\sqrt{H}} \)
To find \( v_0 \) and \( \theta \), we can divide the two equations:
\( \frac{v_0 \sin(\theta)}{v_0 \cos(\theta)} = \frac{\sqrt{\frac{gH}{4}}}{\frac{l \sqrt{g}}{\sqrt{H}}} \)
This simplifies to:
\( \tan(\theta) = \frac{\sqrt{gH}}{2l} \)
From this, we can find \( \theta \) as:
\( \theta = \tan^{-1}\left(\frac{\sqrt{gH}}{2l}\right) \)
Now substituting \( \theta \) back into either equation will allow us to solve for \( v_0 \). Using the vertical motion equation:
\( v_0 = \frac{\sqrt{\frac{gH}{4}}}{\sin(\theta)} \)
By substituting the expression for \( \sin(\theta) \) from the tangent relation, we can find the exact value of \( v_0 \).
Final Thoughts
This problem illustrates the beauty of projectile motion and free fall. By breaking down the motions into their components and using the relationships between them, we can derive the necessary parameters for the thrown body to collide with the falling body at the specified conditions. Understanding these principles is crucial in physics, especially in kinematics and dynamics.
