At what angle with the horizontal should a ball be thrown so that the range `R` is related to the time of flight as `R = 5 T^2` ? `(Take g = 10 ms^(-2))`.

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `90^@`

Explanation

To solve the problem, we need to find the angle \(\theta\) at which a ball should be thrown so that the range \(R\) is related to the time of flight \(T\) by the equation \(R = 5T^2\).

### Step-by-Step Solution:

1. **Understand the equations for range and time of flight:**
- The range \(R\) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
- The time of flight \(T\) is given by:
\[ T = \frac{2u \sin(\theta)}{g} \]

2. **Substitute \(T\) into the range equation:**
- We know from the problem that \(R = 5T^2\).
- Substitute the expression for \(T\) into this equation:
\[ R = 5\left(\frac{2u \sin(\theta)}{g}\right)^2 \]
- Simplifying this gives:
\[ R = 5 \cdot \frac{4u^2 \sin^2(\theta)}{g^2} = \frac{20u^2 \sin^2(\theta)}{g^2} \]

3. **Set the two expressions for \(R\) equal to each other:**
- From the first equation for range, we have:
\[ \frac{u^2 \sin(2\theta)}{g} = \frac{20u^2 \sin^2(\theta)}{g^2} \]
- Cancel \(u^2\) and \(g\) (assuming \(u \neq 0\) and \(g \neq 0\)):
\[ \sin(2\theta) = \frac{20 \sin^2(\theta)}{g} \]

4. **Substitute \(g = 10 \, \text{m/s}^2\):**
- Now, substituting \(g = 10\):
\[ \sin(2\theta) = \frac{20 \sin^2(\theta)}{10} = 2 \sin^2(\theta) \]

5. **Use the double angle identity:**
- Recall that \(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\):
\[ 2 \sin(\theta) \cos(\theta) = 2 \sin^2(\theta) \]
- Dividing both sides by 2 (assuming \(\sin(\theta) \neq 0\)):
\[ \sin(\theta) \cos(\theta) = \sin^2(\theta) \]

6. **Rearranging the equation:**
- This can be rearranged to:
\[ \cos(\theta) = \sin(\theta) \]

7. **Finding the angle:**
- The equation \(\cos(\theta) = \sin(\theta)\) implies:
\[ \tan(\theta) = 1 \]
- Therefore, \(\theta = \tan^{-1}(1) = 45^\circ\).

### Final Answer:
The angle \(\theta\) at which the ball should be thrown is \(45^\circ\).