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

Answer

To solve the problem step by step, we will analyze the motion of the projectile released from the airplane and use the equations of motion.

### Step 1: Understand the problem
The airplane is diving at an angle of \(53^\circ\) with the vertical and releases a projectile from an altitude of 730 m. The projectile hits the ground after 5 seconds. We need to find the speed of the airplane.

### Step 2: Break down the velocity of the airplane
Let the speed of the airplane be \(V_0\). Since the angle with the vertical is \(53^\circ\), we can break the velocity into vertical and horizontal components:
- Vertical component: \(V_{0y} = V_0 \cos(53^\circ)\)
- Horizontal component: \(V_{0x} = V_0 \sin(53^\circ)\)

Using the values of cosine and sine for \(53^\circ\):
- \(\cos(53^\circ) \approx \frac{3}{5}\)
- \(\sin(53^\circ) \approx \frac{4}{5}\)

Thus, we have:
- \(V_{0y} = V_0 \cdot \frac{3}{5}\)
- \(V_{0x} = V_0 \cdot \frac{4}{5}\)

### Step 3: Apply the second equation of motion
For vertical motion, we can use the second equation of motion:
\[ S_y = U_y T + \frac{1}{2} a_y T^2 \]
where:
- \(S_y = 730 \, \text{m}\) (the vertical distance fallen)
- \(U_y = V_{0y} = V_0 \cdot \frac{3}{5}\)
- \(T = 5 \, \text{s}\)
- \(a_y = g = 10 \, \text{m/s}^2\) (acceleration due to gravity)

Substituting these values into the equation gives:
\[ 730 = \left(V_0 \cdot \frac{3}{5}\right) \cdot 5 + \frac{1}{2} \cdot 10 \cdot (5^2) \]

### Step 4: Simplify the equation
Now, simplify the equation:
\[ 730 = 3V_0 + \frac{1}{2} \cdot 10 \cdot 25 \]
\[ 730 = 3V_0 + 125 \]

### Step 5: Solve for \(V_0\)
Rearranging the equation to solve for \(V_0\):
\[ 3V_0 = 730 - 125 \]
\[ 3V_0 = 605 \]
\[ V_0 = \frac{605}{3} \approx 201.67 \, \text{m/s} \]

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