A projectile has a range R and time of flight T. If the range is doubled (by increasing the speed of projection, without changing the angle of projection), the time of flight will become

Options

A. `T// sqrt2`
B. `sqrt2T`
C. `T//2`
D. 2T

Explanation

To solve the problem step-by-step, we will analyze the relationship between the range, time of flight, and the initial speed of projection of a projectile.

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

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

### Step 2: Establish the initial conditions
Let the initial range be \( R = r \) and the time of flight be \( T = t \). Thus, we have:
\[ r = \frac{u^2 \sin 2\theta}{g} \]
\[ t = \frac{2u \sin \theta}{g} \]

### Step 3: Doubling the range
According to the problem, the range is doubled, so the new range \( R' \) becomes:
\[ R' = 2r \]
Substituting into the range formula, we have:
\[ 2r = \frac{u'^2 \sin 2\theta}{g} \]
where \( u' \) is the new speed of projection.

### Step 4: Relate the new speed to the old speed
From the original range equation, we can express \( r \) as:
\[ r = \frac{u^2 \sin 2\theta}{g} \]
Substituting this into the equation for the new range gives:
\[ 2 \left(\frac{u^2 \sin 2\theta}{g}\right) = \frac{u'^2 \sin 2\theta}{g} \]
Cancelling \( \sin 2\theta/g \) from both sides (assuming \( \sin 2\theta \neq 0 \)):
\[ 2u^2 = u'^2 \]
Taking the square root of both sides, we find:
\[ u' = u\sqrt{2} \]

### Step 5: Calculate the new time of flight
Now, we need to find the new time of flight \( T' \) using the new speed \( u' \):
\[ T' = \frac{2u' \sin \theta}{g} \]
Substituting \( u' = u\sqrt{2} \):
\[ T' = \frac{2(u\sqrt{2}) \sin \theta}{g} \]
This simplifies to:
\[ T' = \sqrt{2} \cdot \frac{2u \sin \theta}{g} \]
Recognizing that \( \frac{2u \sin \theta}{g} = T \):
\[ T' = \sqrt{2} \cdot T \]

### Final Answer
Thus, the new time of flight when the range is doubled is:
\[ T' = \sqrt{2} T \]