Two boys simultaneously aim their guns at a bird sitting on a tower. The first boy releases his shot with a speed of 100m/s at an angle of projection of `30^(@)` . The second boy is ahead of the first by a distance of 50 m and releases his shot with a speed of 80m/s. How must he aim his gun so that both the shots hit the bird simultaneously ? What is the distance of the foot of the tower from the two boys and the height of the tower ? With what velocities and when do the two shots hit the bird?

Answer

To solve the problem step by step, we will analyze the projectile motion of both boys' shots and determine the required angle for the second boy to ensure both shots hit the bird simultaneously.

### Step 1: Analyze the first boy's shot
The first boy shoots with a speed of \( u_1 = 100 \, \text{m/s} \) at an angle of \( \theta_1 = 30^\circ \).

**Horizontal and Vertical Components:**
- Horizontal component:
\[ u_{1x} = u_1 \cos(\theta_1) = 100 \cos(30^\circ) = 100 \times \frac{\sqrt{3}}{2} = 50\sqrt{3} \, \text{m/s} \]
- Vertical component:
\[ u_{1y} = u_1 \sin(\theta_1) = 100 \sin(30^\circ) = 100 \times \frac{1}{2} = 50 \, \text{m/s} \]

### Step 2: Determine the time of flight for the first boy
The time of flight \( T \) can be calculated using the formula for the vertical motion:
\[ h = u_{1y} T - \frac{1}{2} g T^2 \]
where \( h \) is the height of the tower, and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 3: Analyze the second boy's shot
The second boy is 50 m ahead and shoots with a speed of \( u_2 = 80 \, \text{m/s} \) at an angle \( \theta_2 \).

**Horizontal and Vertical Components:**
- Horizontal component:
\[ u_{2x} = u_2 \cos(\theta_2) = 80 \cos(\theta_2) \, \text{m/s} \]
- Vertical component:
\[ u_{2y} = u_2 \sin(\theta_2) = 80 \sin(\theta_2) \, \text{m/s} \]

### Step 4: Set up the equations for horizontal motion
Both shots must hit the bird at the same time \( T \). The horizontal distance traveled by both boys can be expressed as:
- For the first boy:
\[ \text{Distance} = u_{1x} T = 50\sqrt{3} T \]
- For the second boy (who is 50 m ahead):
\[ \text{Distance} = 50 + u_{2x} T = 50 + 80 \cos(\theta_2) T \]

Setting the distances equal gives:
\[ 50\sqrt{3} T = 50 + 80 \cos(\theta_2) T \]

### Step 5: Solve for \( \theta_2 \)
Rearranging the equation:
\[ 50\sqrt{3} T - 80 \cos(\theta_2) T = 50 \]
Factoring out \( T \):
\[ T (50\sqrt{3} - 80 \cos(\theta_2)) = 50 \]
Thus,
\[ T = \frac{50}{50\sqrt{3} - 80 \cos(\theta_2)} \]

### Step 6: Set up the equations for vertical motion
For the vertical motion, both projectiles must reach the same height \( h \):
- For the first boy:
\[ h = 50T - \frac{1}{2} g T^2 \]
- For the second boy:
\[ h = 80 \sin(\theta_2) T - \frac{1}{2} g T^2 \]

Setting these equal gives:
\[ 50T - \frac{1}{2} g T^2 = 80 \sin(\theta_2) T - \frac{1}{2} g T^2 \]
This simplifies to:
\[ 50T = 80 \sin(\theta_2) T \]
Thus,
\[ \sin(\theta_2) = \frac{50}{80} = \frac{5}{8} \]

### Step 7: Find \( \theta_2 \)
Using the inverse sine function:
\[ \theta_2 = \sin^{-1}\left(\frac{5}{8}\right) \]

### Step 8: Calculate the height of the tower and the distance from the boys
Substituting \( T \) back into the height equation:
\[ h = 50T - \frac{1}{2} g T^2 \]
Using \( g \approx 9.81 \, \text{m/s}^2 \) and the value of \( T \) from earlier, we can find \( h \).

### Step 9: Determine the velocities when they hit the bird
The velocities can be calculated using:
- For the first boy:
\[ v_{1} = \sqrt{u_{1x}^2 + (u_{1y} - gT)^2} \]
- For the second boy:
\[ v_{2} = \sqrt{u_{2x}^2 + (u_{2y} - gT)^2} \]

### Summary
1. The second boy must aim at an angle \( \theta_2 = \sin^{-1}\left(\frac{5}{8}\right) \).
2. The height of the tower and the distance from the boys can be calculated using the derived equations.