A person observes a bird on a tree 39.6 m high and at a distance of 59.2 m. With what velocity the person should throw an arrow at an angle of `45^(@)` so that it may hit the bird ?

Answer

To solve the problem of determining the velocity at which a person should throw an arrow at an angle of 45 degrees to hit a bird perched on a tree 39.6 m high and 59.2 m away, we can follow these steps:

### Step 1: Understand the problem
We need to find the initial velocity \( v_0 \) required to hit a target (the bird) that is at a certain height (39.6 m) and horizontal distance (59.2 m) from the thrower. The angle of projection is given as \( 45^\circ \).

### Step 2: Break down the motion
The motion can be analyzed in two dimensions: horizontal (x) and vertical (y).

- **Horizontal motion**:
\[ x = v_0 \cos(\theta) \cdot t \]
where \( x = 59.2 \, \text{m} \) and \( \theta = 45^\circ \).

- **Vertical motion**:
\[ y = v_0 \sin(\theta) \cdot t - \frac{1}{2} g t^2 \]
where \( y = 39.6 \, \text{m} \) and \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).

### Step 3: Substitute the angle
Since \( \theta = 45^\circ \), we have:
\[ \cos(45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.707 \]

### Step 4: Set up the equations
From the horizontal motion:
\[ 59.2 = v_0 \cdot 0.707 \cdot t \quad \text{(1)} \]

From the vertical motion:
\[ 39.6 = v_0 \cdot 0.707 \cdot t - \frac{1}{2} \cdot 9.81 \cdot t^2 \quad \text{(2)} \]

### Step 5: Solve for \( t \) from equation (1)
Rearranging equation (1) gives:
\[ t = \frac{59.2}{v_0 \cdot 0.707} \quad \text{(3)} \]

### Step 6: Substitute \( t \) into equation (2)
Substituting equation (3) into equation (2):
\[ 39.6 = v_0 \cdot 0.707 \cdot \left(\frac{59.2}{v_0 \cdot 0.707}\right) - \frac{1}{2} \cdot 9.81 \cdot \left(\frac{59.2}{v_0 \cdot 0.707}\right)^2 \]

### Step 7: Simplify the equation
This simplifies to:
\[ 39.6 = 59.2 - \frac{1}{2} \cdot 9.81 \cdot \frac{59.2^2}{(v_0 \cdot 0.707)^2} \]

### Step 8: Rearranging and solving for \( v_0 \)
Rearranging gives:
\[ \frac{1}{2} \cdot 9.81 \cdot \frac{59.2^2}{(v_0 \cdot 0.707)^2} = 59.2 - 39.6 \]
\[ \frac{1}{2} \cdot 9.81 \cdot \frac{59.2^2}{(v_0 \cdot 0.707)^2} = 19.6 \]

Multiplying both sides by \( 2(v_0 \cdot 0.707)^2 \):
\[ 9.81 \cdot 59.2^2 = 19.6 \cdot 2(v_0 \cdot 0.707)^2 \]
\[ v_0^2 = \frac{9.81 \cdot 59.2^2}{39.2 \cdot 0.707^2} \]

### Step 9: Calculate \( v_0 \)
Calculating the right-hand side:
\[ v_0^2 = \frac{9.81 \cdot 3490.24}{39.2 \cdot 0.4998} \]
\[ v_0^2 = \frac{34229.78}{19.592} \]
\[ v_0^2 \approx 1745.15 \]
\[ v_0 \approx \sqrt{1745.15} \approx 41.86 \, \text{m/s} \]

### Final Answer
The initial velocity \( v_0 \) required to hit the bird is approximately **41.86 m/s**.