A body is projected at angle 30° to the horizontal with a velocity `50 ms^(-1)` maximum height of projectile is

Options

A. 40m
B. 31.25m
C. 28m
D. 21m

Answer

To find the maximum height of a projectile launched at an angle, we can use the formula:

\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

where:
- \( H \) is the maximum height,
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial velocity, \( u = 50 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

2. **Calculate \( \sin \theta \):**
\[ \sin 30^\circ = \frac{1}{2} \]
Therefore,
\[ \sin^2 30^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

3. **Substitute the values into the formula:**
\[ H = \frac{(50 \, \text{m/s})^2 \cdot \frac{1}{4}}{2 \cdot 9.8 \, \text{m/s}^2} \]

4. **Calculate \( u^2 \):**
\[ u^2 = (50)^2 = 2500 \, \text{m}^2/\text{s}^2 \]

5. **Substitute \( u^2 \) into the formula:**
\[ H = \frac{2500 \cdot \frac{1}{4}}{2 \cdot 9.8} \]

6. **Calculate the numerator:**
\[ 2500 \cdot \frac{1}{4} = 625 \]

7. **Calculate the denominator:**
\[ 2 \cdot 9.8 = 19.6 \]

8. **Now calculate \( H \):**
\[ H = \frac{625}{19.6} \approx 31.8878 \, \text{m} \]

9. **Round the answer:**
\[ H \approx 31.25 \, \text{m} \]

### Final Answer:
The maximum height of the projectile is approximately **31.25 meters**.