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 ms6-2)`.

Options

A. (a)`30^@`
B. (b)`45^@`
C. (c)`60^@`
D. (d)`90^@`

Explanation

To solve the problem of finding the angle at which a ball should be thrown so that the range \( R \) is related to the time of flight \( T \) as \( R = 5T^2 \), we can follow these steps:

### Step 1: Understand the relationship between range and time of flight
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{V^2 \sin 2\theta}{g} \]
where \( V \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2V \sin \theta}{g} \]

### Step 2: Express \( R \) in terms of \( T \)
From the problem, we have:
\[ R = 5T^2 \]
We can substitute \( T \) from the time of flight formula into this equation. First, we solve for \( T \):
\[ T = \frac{2V \sin \theta}{g} \]
Now, substituting \( T \) into \( R = 5T^2 \):
\[ R = 5\left(\frac{2V \sin \theta}{g}\right)^2 \]
This simplifies to:
\[ R = 5 \cdot \frac{4V^2 \sin^2 \theta}{g^2} = \frac{20V^2 \sin^2 \theta}{g^2} \]

### Step 3: Set the two expressions for \( R \) equal
Now we have two expressions for \( R \):
1. \( R = \frac{V^2 \sin 2\theta}{g} \)
2. \( R = \frac{20V^2 \sin^2 \theta}{g^2} \)

Setting them equal to each other:
\[ \frac{V^2 \sin 2\theta}{g} = \frac{20V^2 \sin^2 \theta}{g^2} \]

### Step 4: Simplify the equation
We can cancel \( V^2 \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{\sin 2\theta}{g} = \frac{20 \sin^2 \theta}{g^2} \]
Multiplying both sides by \( g^2 \):
\[ g \sin 2\theta = 20 \sin^2 \theta \]

### Step 5: Use the identity for \( \sin 2\theta \)
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ g(2 \sin \theta \cos \theta) = 20 \sin^2 \theta \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ 10(2 \sin \theta \cos \theta) = 20 \sin^2 \theta \]
This simplifies to:
\[ 20 \sin \theta \cos \theta = 20 \sin^2 \theta \]
Dividing both sides by 20:
\[ \sin \theta \cos \theta = \sin^2 \theta \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ \sin \theta \cos \theta - \sin^2 \theta = 0 \]
Factoring out \( \sin \theta \):
\[ \sin \theta (\cos \theta - \sin \theta) = 0 \]

### Step 7: Solving for \( \theta \)
This gives us two cases:
1. \( \sin \theta = 0 \) (which corresponds to \( \theta = 0^\circ \) or \( 180^\circ \), not valid for projectile motion)
2. \( \cos \theta - \sin \theta = 0 \) which leads to \( \cos \theta = \sin \theta \)

This implies:
\[ \tan \theta = 1 \]
Thus, \( \theta = 45^\circ \).

### Conclusion
The angle at which the ball should be thrown is \( \theta = 45^\circ \).

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