The maximum height attained by a ball projected with speed ` 20 ms^(-1)` at an angle `45^(@)` with the horizontal is [take g = ` 10 ms^(-2)`]

Options

A. 40 m
B. 20 m
C. 10 m
D. 30 m

Answer

To find the maximum height attained by a ball projected with a speed of \( 20 \, \text{m/s} \) at an angle of \( 45^\circ \) with the horizontal, we can use the formula for maximum height in projectile motion:

\[ h_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g} \]

Where:
- \( h_{\text{max}} \) is the maximum height,
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step 1: Identify the values
- Given \( u = 20 \, \text{m/s} \)
- Given \( \theta = 45^\circ \)
- Given \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate \( \sin^2 \theta \)
For \( \theta = 45^\circ \):
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \]
Thus,
\[ \sin^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

### Step 3: Substitute the values into the formula
Now, substituting the values into the formula:
\[ h_{\text{max}} = \frac{(20)^2 \cdot \frac{1}{2}}{2 \cdot 10} \]

### Step 4: Simplify the expression
Calculating the numerator:
\[ (20)^2 = 400 \]
Thus,
\[ h_{\text{max}} = \frac{400 \cdot \frac{1}{2}}{20} \]
This simplifies to:
\[ h_{\text{max}} = \frac{200}{20} = 10 \, \text{m} \]

### Conclusion
The maximum height attained by the ball is:
\[ \boxed{10 \, \text{m}} \]