A particle is projected from ground with some initial velocity making an angle of `45^(@)` with the horizontal. It reaches a height of `7.5 m` above the ground while it travels a horizontal distance of `10 m` from the point of projection. The initial speed of the projection is

Options

A. `5 m//s`
B. `10 m//s`
C. `20 m//s`
D. `40 m//s`

Explanation

To solve the problem, we need to find the initial speed of a particle projected at an angle of \(45^\circ\) with respect to the horizontal, which reaches a height of \(7.5 \, m\) and travels a horizontal distance of \(10 \, m\).

### Step-by-Step Solution:

1. **Understanding the Problem**:
The particle is projected at an angle of \(45^\circ\). This means that the initial velocity can be broken down into horizontal and vertical components:
\[ u_x = u \cos(45^\circ) = \frac{u}{\sqrt{2}} \]
\[ u_y = u \sin(45^\circ) = \frac{u}{\sqrt{2}} \]

2. **Vertical Motion**:
The maximum height \(h\) reached by the projectile is given as \(7.5 \, m\). The formula for maximum height in projectile motion is:
\[ h = \frac{u_y^2}{2g} \]
Substituting \(u_y\):
\[ 7.5 = \frac{\left(\frac{u}{\sqrt{2}}\right)^2}{2g} \]
Simplifying this:
\[ 7.5 = \frac{u^2}{2 \cdot 2g} = \frac{u^2}{4g} \]
Rearranging gives:
\[ u^2 = 30g \]

3. **Horizontal Motion**:
The horizontal distance \(R\) traveled by the projectile is given as \(10 \, m\). The time of flight \(t\) can be calculated using the horizontal motion:
\[ R = u_x \cdot t \]
Since \(u_x = \frac{u}{\sqrt{2}}\):
\[ 10 = \frac{u}{\sqrt{2}} \cdot t \]
Rearranging gives:
\[ t = \frac{10\sqrt{2}}{u} \]

4. **Finding Time of Flight**:
The time of flight for a projectile launched at an angle can also be expressed in terms of its vertical motion. The total time of flight \(t\) can be calculated using:
\[ t = \frac{2u_y}{g} = \frac{2 \cdot \frac{u}{\sqrt{2}}}{g} = \frac{\sqrt{2}u}{g} \]

5. **Equating the Two Expressions for Time**:
Now we have two expressions for time \(t\):
\[ \frac{10\sqrt{2}}{u} = \frac{\sqrt{2}u}{g} \]
Cross-multiplying gives:
\[ 10g = u^2 \]

6. **Substituting for \(g\)**:
We can substitute \(g \approx 9.8 \, m/s^2\):
\[ u^2 = 10 \cdot 9.8 = 98 \]
Thus:
\[ u = \sqrt{98} \approx 9.9 \, m/s \]

### Final Answer:
The initial speed of the projection is approximately \(9.9 \, m/s\).