If a stone is to hit at a point which is at a horizontal distance 100 m away and at a hight 50 m above the point from where the stone starts, then what is the value of initial speed u if the stone is launched at an angle `45^(@)`?

Options

A. `10sqrt2 m//s`
B. `10sqrt5m//s`
C. `20sqrt5m//s`
D. `20sqrt(10)m//s`

Answer

To solve the problem of finding the initial speed \( u \) required for a stone to hit a target at a horizontal distance of 100 m and a height of 50 m when launched at an angle of \( 45^\circ \), we can follow these steps:

### Step 1: Define the known variables
- Horizontal distance (\( x \)): 100 m
- Height (\( y \)): 50 m
- Launch angle (\( \theta \)): \( 45^\circ \)
- Acceleration due to gravity (\( g \)): approximately \( 10 \, \text{m/s}^2 \)

### Step 2: Break down the initial velocity into components
The initial velocity \( u \) can be broken down into horizontal and vertical components:
- Horizontal component (\( u_x \)):
\[ u_x = u \cos \theta \]
- Vertical component (\( u_y \)):
\[ u_y = u \sin \theta \]

Since \( \theta = 45^\circ \), we have:
\[ \cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}} \]
Thus,
\[ u_x = u \cdot \frac{1}{\sqrt{2}}, \quad u_y = u \cdot \frac{1}{\sqrt{2}} \]

### Step 3: Find the time of flight
The horizontal distance covered is given by:
\[ x = u_x \cdot t \implies t = \frac{x}{u_x} = \frac{100}{u \cdot \frac{1}{\sqrt{2}}} = \frac{100 \sqrt{2}}{u} \]

### Step 4: Use the vertical motion equation
The vertical motion can be described by the equation:
\[ y = u_y \cdot t - \frac{1}{2} g t^2 \]
Substituting for \( y \), \( u_y \), and \( t \):
\[ 50 = u \cdot \frac{1}{\sqrt{2}} \cdot \left( \frac{100 \sqrt{2}}{u} \right) - \frac{1}{2} g \left( \frac{100 \sqrt{2}}{u} \right)^2 \]

### Step 5: Simplify the equation
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ 50 = 100 - \frac{1}{2} \cdot 10 \cdot \left( \frac{100 \sqrt{2}}{u} \right)^2 \]
\[ 50 = 100 - 5 \cdot \frac{20000}{u^2} \]
\[ 50 = 100 - \frac{100000}{u^2} \]
\[ \frac{100000}{u^2} = 50 \]
\[ u^2 = \frac{100000}{50} = 2000 \]
\[ u = \sqrt{2000} = 44.72 \, \text{m/s} \]

### Final Answer
The initial speed \( u \) required for the stone to hit the target is approximately \( 44.72 \, \text{m/s} \).