An airplane, diving at an angle of `53.0^@` with the vertical releases a projectile at an altitude of 730 m. The projectile hits the ground 5.00 s after being released. What is the speed of the aircraft?

Options

A. `282ms^(-1)`
B. `202ms(-1)`
C. `182ms(-1)`
D. `102ms(-1)`

Explanation

To solve the problem, we need to determine the speed of the aircraft diving at an angle of \(53.0^\circ\) with the vertical when it releases a projectile from an altitude of 730 m, which hits the ground after 5.00 seconds.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The projectile is released from a height of 730 m and takes 5 seconds to hit the ground. We will analyze the vertical motion of the projectile to find the initial vertical component of the velocity.

2. **Using the Second Equation of Motion**:
The second equation of motion in the vertical direction is given by:
\[ h = u_y t + \frac{1}{2} g t^2 \]
where:
- \(h\) is the height (730 m),
- \(u_y\) is the initial vertical component of the velocity,
- \(t\) is the time (5 s),
- \(g\) is the acceleration due to gravity (approximately \(10 \, \text{m/s}^2\)).

3. **Setting Up the Equation**:
Since we are taking the downward direction as positive, we can rearrange the equation:
\[ 730 = u_y \cdot 5 + \frac{1}{2} \cdot 10 \cdot (5^2) \]
Simplifying the term with gravity:
\[ 730 = u_y \cdot 5 + 125 \]

4. **Solving for \(u_y\)**:
Rearranging the equation to solve for \(u_y\):
\[ u_y \cdot 5 = 730 - 125 \]
\[ u_y \cdot 5 = 605 \]
\[ u_y = \frac{605}{5} = 121 \, \text{m/s} \]

5. **Finding the Speed of the Aircraft**:
The speed of the aircraft \(u\) can be found using the angle of the dive. The vertical component of the speed \(u_y\) is related to the total speed \(u\) and the angle \(\theta\) (which is \(53^\circ\) from the vertical) by:
\[ u_y = u \cdot \cos(53^\circ) \]
Since \(\cos(53^\circ) = \frac{3}{5}\):
\[ 121 = u \cdot \frac{3}{5} \]
Rearranging gives:
\[ u = \frac{121 \cdot 5}{3} = \frac{605}{3} \approx 201.67 \, \text{m/s} \]

6. **Final Answer**:
The speed of the aircraft is approximately \(202 \, \text{m/s}\).