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

Options

A. `50(sqrt3-1)ms^(-1)`
B. `50(sqrt2-1)ms^(-1)`
C. `70(sqrt3-1)ms^(-1)`
D. `80(sqrt3-1)ms^(-1)`

Answer

To solve the problem, we need to determine the height of the tower and the speed with which the ball was projected. We will use the principles of projectile motion.

### Step 1: Analyze the given information
- The ball is projected at an angle of \(30^\circ\) above the horizontal.
- It strikes the ground after \(5\) seconds at an angle of \(45^\circ\) with the horizontal.

### Step 2: Break down the initial velocity
Let the initial speed of projection be \(u\).
- The horizontal component of the initial velocity is:
\[ u_x = u \cos(30^\circ) = u \cdot \frac{\sqrt{3}}{2} \]
- The vertical component of the initial velocity is:
\[ u_y = u \sin(30^\circ) = u \cdot \frac{1}{2} \]

### Step 3: Analyze the final velocity components
At the time of impact, the ball strikes the ground at an angle of \(45^\circ\). This means that the horizontal and vertical components of the final velocity (\(v_x\) and \(v_y\)) are equal:
\[ v_x = v_y \]
Since there is no horizontal acceleration, we have:
\[ v_x = u_x = u \cdot \frac{\sqrt{3}}{2} \]

### Step 4: Determine the vertical component of the final velocity
Using the equations of motion, the vertical component of the final velocity can be expressed as:
\[ v_y = u_y - g t \]
Substituting the values:
\[ v_y = u \cdot \frac{1}{2} - 10 \cdot 5 \]
\[ v_y = u \cdot \frac{1}{2} - 50 \]

### Step 5: Set the components equal
Since \(v_x = v_y\), we can set the equations equal:
\[ u \cdot \frac{\sqrt{3}}{2} = u \cdot \frac{1}{2} - 50 \]

### Step 6: Solve for \(u\)
Rearranging the equation:
\[ u \cdot \frac{\sqrt{3}}{2} - u \cdot \frac{1}{2} = -50 \]
Factoring out \(u\):
\[ u \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) = -50 \]
\[ u \cdot \frac{\sqrt{3} - 1}{2} = 50 \]
\[ u = \frac{100}{\sqrt{3} - 1} \]
To rationalize the denominator:
\[ u = \frac{100(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{100(\sqrt{3} + 1)}{2} = 50(\sqrt{3} + 1) \]

### Step 7: Calculate the height of the tower
Using the equation of motion for vertical displacement:
\[ h = u_y t + \frac{1}{2}(-g)t^2 \]
Substituting the values:
\[ h = \left(u \cdot \frac{1}{2}\right) \cdot 5 - \frac{1}{2} \cdot 10 \cdot 5^2 \]
\[ h = \left(50(\sqrt{3} + 1) \cdot \frac{1}{2}\right) \cdot 5 - \frac{1}{2} \cdot 10 \cdot 25 \]
\[ h = 125(\sqrt{3} + 1) - 125 \]
\[ h = 125\sqrt{3} \]

### Final Answers
- The speed with which the ball was projected is \(50(\sqrt{3} + 1) \, \text{m/s}\).
- The height of the tower is \(125\sqrt{3} \, \text{m}\).