To solve this problem, we need to analyze the motion of both the ball and the bird. The ball is thrown from the ground at an angle, while the bird starts from a height and moves horizontally. The key is to find the distance on the ground where the ball strikes, given that it just touches the bird at one point during its flight. Let's break this down step by step.
Understanding the Motion of the Ball
The ball is projected from the ground with an initial velocity \( v \) at an angle \( \theta \). The horizontal and vertical components of the ball's velocity can be expressed as:
- Horizontal component: \( v_x = v \cos(\theta) \)
- Vertical component: \( v_y = v \sin(\theta) \)
The equations of motion for the ball can be described as follows:
- Horizontal distance covered: \( x = v_x t = (v \cos(\theta)) t \)
- Vertical distance covered: \( y = v_y t - \frac{1}{2} g t^2 = (v \sin(\theta)) t - \frac{1}{2} g t^2 \)
Analyzing the Bird's Motion
The bird starts at a height \( h \) and moves horizontally with a constant speed \( u \). The horizontal distance covered by the bird after time \( t \) is:
- Distance: \( x_b = u t \)
- Height remains constant: \( y_b = h \)
Finding the Point of Intersection
For the ball to just touch the bird, their horizontal distances must be equal at the same time \( t \), and the vertical position of the ball must equal the height of the bird:
- From the horizontal motion: \( v \cos(\theta) t = u t \)
- From the vertical motion: \( (v \sin(\theta)) t - \frac{1}{2} g t^2 = h \)
From the first equation, we can simplify to find the relationship between the velocities:
Since \( t \) cannot be zero, we can divide both sides by \( t \):
Thus, we have:
\( v \cos(\theta) = u \)
Substituting into the Vertical Motion Equation
Now, substituting \( v \) from the first equation into the vertical motion equation:
Let \( v = \frac{u}{\cos(\theta)} \), then:
\( \left(\frac{u}{\cos(\theta)} \sin(\theta)\right) t - \frac{1}{2} g t^2 = h \)
This simplifies to:
\( \frac{u \tan(\theta)}{\cos(\theta)} t - \frac{1}{2} g t^2 = h \)
Solving for Time \( t \)
Rearranging gives us a quadratic equation in \( t \):
\( -\frac{1}{2} g t^2 + \frac{u \tan(\theta)}{\cos(\theta)} t - h = 0 \)
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where:
- \( a = -\frac{1}{2} g \)
- \( b = \frac{u \tan(\theta)}{\cos(\theta)} \)
- \( c = -h \)
Finding the Horizontal Distance
Once we find \( t \), we can substitute it back into the horizontal distance equation:
\( x = v \cos(\theta) t \)
Substituting \( v \) gives:
\( x = u t \)
Final Expression for Distance
After solving the quadratic equation and substituting back, we arrive at the distance where the ball strikes the ground:
\( x = 2u \sqrt{\frac{2h}{g}} \)
This result shows that the distance on the ground where the ball strikes is directly proportional to the horizontal speed of the bird and the square root of the height from which the bird starts. This relationship highlights the interplay between vertical and horizontal motions in projectile dynamics.
