A Bomber flying upward at an angle of `53^(@)` with the vertical releases a bomb at an altitude of 800 m. The bomb strikes the ground 20 s after its release. Find : [Given `sin 53^(@)=0.8, g=10 m//s^(2)`]
(i) The velocity of the bomber at the time of release of the bomb.
(ii) The maximum height attained by the bomb.
(iii) The horizontal distance tranvelled by the bomb before it strikes the ground
(iv) The velocity (magnitude & direction) of the bomb just when it strikes the ground.

Answer

To solve the problem step by step, we will break it down into the required parts as specified in the question.

### Given Data:
- Angle with the vertical, θ = 53°
- Height of release, h = 800 m
- Time of flight, t = 20 s
- Acceleration due to gravity, g = 10 m/s²
- sin(53°) = 0.8
- cos(53°) = 0.6 (derived from sin² + cos² = 1)

### Step 1: Finding the Vertical Component of the Initial Velocity (u_y)

Using the vertical motion equation:
\[ \Delta y = u_y t - \frac{1}{2} g t^2 \]
Here, \(\Delta y = -800\) m (since the bomb falls down), \(t = 20\) s, and \(g = 10\) m/s².

Substituting the values:
\[ -800 = u_y (20) - \frac{1}{2} (10) (20^2) \]
\[ -800 = 20u_y - 1000 \]
\[ 20u_y = 200 \]
\[ u_y = 10 \text{ m/s} \]

### Step 2: Finding the Horizontal Component of the Initial Velocity (u_x)

Using the relation between the components:
\[ \tan(θ) = \frac{u_y}{u_x} \]
From the angle given, \(\tan(53°) = \frac{3}{4}\). Thus,
\[ \frac{3}{4} = \frac{10}{u_x} \]
Cross-multiplying gives:
\[ 3u_x = 40 \implies u_x = \frac{40}{3} \approx 13.33 \text{ m/s} \]

### Step 3: Finding the Magnitude of the Initial Velocity (u)

The magnitude of the initial velocity can be found using the Pythagorean theorem:
\[ u = \sqrt{u_x^2 + u_y^2} \]
Substituting the values:
\[ u = \sqrt{(13.33)^2 + (10)^2} = \sqrt{177.69 + 100} = \sqrt{277.69} \approx 16.67 \text{ m/s} \]

### Step 4: Finding the Maximum Height Attained by the Bomb

The maximum height (H_max) from the point of release can be calculated using:
\[ H_{max} = \frac{u_y^2}{2g} \]
Substituting the values:
\[ H_{max} = \frac{(10)^2}{2 \times 10} = \frac{100}{20} = 5 \text{ m} \]
Thus, the total height from the ground:
\[ H_{total} = 800 + H_{max} = 800 + 5 = 805 \text{ m} \]

### Step 5: Finding the Horizontal Distance Traveled by the Bomb

The horizontal distance (R) can be calculated using:
\[ R = u_x \cdot t \]
Substituting the values:
\[ R = 13.33 \cdot 20 = 266.67 \text{ m} \]

### Step 6: Finding the Velocity of the Bomb Just Before It Strikes the Ground

The horizontal component remains constant:
\[ u_x = 13.33 \text{ m/s} \]
The vertical component just before striking the ground can be calculated using:
\[ v_y = u_y - g \cdot t \]
Substituting the values:
\[ v_y = 10 - 10 \cdot 20 = 10 - 200 = -190 \text{ m/s} \]

### Step 7: Finding the Magnitude and Direction of the Velocity Just Before Impact

The resultant velocity can be calculated using:
\[ v = \sqrt{u_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(13.33)^2 + (-190)^2} = \sqrt{177.69 + 36100} = \sqrt{36277.69} \approx 190.5 \text{ m/s} \]

The direction can be found using:
\[ \tan(\theta) = \frac{v_y}{u_x} \]
\[ \theta = \tan^{-1}\left(\frac{-190}{13.33}\right) \approx -86.3° \]

### Summary of Results:
1. The velocity of the bomber at the time of release of the bomb: **16.67 m/s**
2. The maximum height attained by the bomb: **805 m**
3. The horizontal distance traveled by the bomb before it strikes the ground: **266.67 m**
4. The velocity (magnitude & direction) of the bomb just when it strikes the ground: **190.5 m/s at -86.3°**