A ball is thrown at an angle `theta` with the horizontal and the range is maximum. The value of `tantheta` is:-

Options

A. 4
B. 2
C. 1
D. 0.5

Answer

To solve the problem step by step, we need to determine the angle `theta` at which the range of a projectile is maximized and then find the value of `tan(theta)` at that angle.

### Step-by-Step Solution:

1. **Understanding the Range of a Projectile**:
The range \( R \) of a projectile launched with an initial velocity \( u \) at an angle \( \theta \) with the horizontal is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

2. **Maximizing the Range**:
To maximize the range, we need to maximize the term \( \sin(2\theta) \). The sine function reaches its maximum value of 1.

3. **Finding the Angle**:
The maximum value of \( \sin(2\theta) \) occurs when:
\[ 2\theta = 90^\circ \]
This implies:
\[ \theta = 45^\circ \]

4. **Calculating \( \tan(\theta) \)**:
Now we need to find \( \tan(\theta) \) when \( \theta = 45^\circ \):
\[ \tan(45^\circ) = 1 \]

5. **Conclusion**:
Therefore, the value of \( \tan(\theta) \) when the range of the projectile is maximum is:
\[ \tan(\theta) = 1 \]

### Final Answer:
The value of \( \tan(\theta) \) is \( 1 \).

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