If the initial velocity of a projectile be doubled, keeping the angle of projection same, the maximum height reached by it will

Options

A. remain the same
B. be doubled
C. become four times
D. be halved

Explanation

To solve the problem, we need to analyze how the maximum height of a projectile changes when the initial velocity is doubled while keeping the angle of projection constant.

### Step-by-Step Solution:

1. **Understand the Formula for Maximum Height**:
The maximum height \( h \) reached by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where:
- \( u \) = initial velocity of the projectile,
- \( \theta \) = angle of projection,
- \( g \) = acceleration due to gravity.

2. **Identify the Variables**:
In this scenario, we are told that the initial velocity \( u \) is doubled. Therefore, if the initial velocity is \( u_1 \), the new initial velocity \( u_2 \) will be:
\[ u_2 = 2u_1 \]

3. **Substitute the New Velocity into the Formula**:
We can express the new maximum height \( h_2 \) with the new initial velocity:
\[ h_2 = \frac{(u_2)^2 \sin^2 \theta}{2g} = \frac{(2u_1)^2 \sin^2 \theta}{2g} \]

4. **Simplify the Expression**:
Now, simplify the expression for \( h_2 \):
\[ h_2 = \frac{4u_1^2 \sin^2 \theta}{2g} = 2 \cdot \frac{u_1^2 \sin^2 \theta}{2g} \]
This can be rewritten as:
\[ h_2 = 4 \cdot \left(\frac{u_1^2 \sin^2 \theta}{2g}\right) = 4h_1 \]
where \( h_1 \) is the maximum height reached with the initial velocity \( u_1 \).

5. **Conclusion**:
From the above calculations, we find that when the initial velocity is doubled, the maximum height becomes:
\[ h_2 = 4h_1 \]
Therefore, the maximum height reached by the projectile will be **4 times** the initial height.

### Final Answer:
The maximum height reached by the projectile will be **4 times** the initial height.

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