A ball is projected at an angle of `30^(@)` above with the horizontal from the top of a tower ans strikes the ground in `5sec` at angle of `45^(@)` with the horizontal. Find the hieght of the lower and the speede with which it was projected.

Explanation

To solve the problem, we will break it down into steps. We need to find the height of the tower and the speed with which the ball was projected.

### Step 1: Analyze the motion
The ball is projected at an angle of \(30^\circ\) above the horizontal. We can break down the initial velocity \(u\) into its horizontal and vertical components:
- Horizontal component: \(u_x = u \cos(30^\circ)\)
- Vertical component: \(u_y = u \sin(30^\circ)\)

### Step 2: Determine the final velocity components
The ball strikes the ground at an angle of \(45^\circ\) with the horizontal after \(5\) seconds. At this point, the horizontal and vertical components of the final velocity \(V\) are equal:
- \(V_x = V_y\)

Since the horizontal component remains constant (no horizontal acceleration), we have:
- \(V_x = u_x = u \cos(30^\circ)\)

The vertical component can be calculated using the formula:
\[ V_y = u_y - g t \]
Where \(g\) is the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)) and \(t = 5 \, \text{s}\).

### Step 3: Set up the equations
From the angle of projection:
1. \(V_x = V_y\)
2. \(V_y = u_y - g t\)

Substituting \(u_y\):
\[ V_y = u \sin(30^\circ) - g \cdot 5 \]

Since \(V_x = V_y\):
\[ u \cos(30^\circ) = u \sin(30^\circ) - g \cdot 5 \]

### Step 4: Solve for initial velocity \(u\)
Substituting the values of \(\sin(30^\circ) = \frac{1}{2}\) and \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\):
\[ u \cdot \frac{\sqrt{3}}{2} = u \cdot \frac{1}{2} - 9.81 \cdot 5 \]

Rearranging gives:
\[ u \cdot \frac{\sqrt{3}}{2} - u \cdot \frac{1}{2} = -49.05 \]
\[ u \left( \frac{\sqrt{3}}{2} - \frac{1}{2} \right) = -49.05 \]

Calculating the left side:
\[ \frac{\sqrt{3} - 1}{2} \approx 0.366 \]
Thus:
\[ u \cdot 0.366 = 49.05 \]
\[ u = \frac{49.05}{0.366} \approx 134.3 \, \text{m/s} \]

### Step 5: Calculate the height of the tower
Now that we have \(u\), we can find the height of the tower using the vertical motion equation:
\[ h = u_y t - \frac{1}{2} g t^2 \]
Where \(u_y = u \sin(30^\circ) = 134.3 \cdot \frac{1}{2} = 67.15 \, \text{m/s}\):
\[ h = 67.15 \cdot 5 - \frac{1}{2} \cdot 9.81 \cdot 5^2 \]
\[ h = 335.75 - \frac{1}{2} \cdot 9.81 \cdot 25 \]
\[ h = 335.75 - 122.625 \]
\[ h \approx 213.125 \, \text{m} \]

### Final Answers
- **Height of the tower**: \(213.125 \, \text{m}\)
- **Speed with which the ball was projected**: \(134.3 \, \text{m/s}\)