The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is `20 m//s^(-1)`,its time of flight is `(g=10 m//s^(2))`

Options

A. `2s`
B. `2sqrt2s`
C. `1/2s`
D. `1/sqrt2s`

Explanation

To solve the problem, we need to find the time of flight of a projectile given that its initial velocity \( u = 20 \, \text{m/s} \) and the acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Understanding the Energy Relationship**:
At the maximum height of the projectile, the potential energy (PE) is equal to the kinetic energy (KE). The equations for PE and KE are:
\[ PE = mgh \]
\[ KE = \frac{1}{2} mv^2 \]
Since \( PE = KE \), we can equate them:
\[ mgh = \frac{1}{2} mv^2 \]

2. **Canceling Mass**:
Since mass \( m \) appears on both sides of the equation, we can cancel it out:
\[ gh = \frac{1}{2} v^2 \]

3. **Substituting Known Values**:
We know \( g = 10 \, \text{m/s}^2 \) and \( v = u \cos \theta \) where \( u = 20 \, \text{m/s} \):
\[ 10h = \frac{1}{2} (20 \cos \theta)^2 \]
Simplifying the right side:
\[ 10h = \frac{1}{2} \cdot 400 \cos^2 \theta \]
\[ 10h = 200 \cos^2 \theta \]
Thus,
\[ h = 20 \cos^2 \theta \quad \text{(Equation 1)} \]

4. **Finding Maximum Height**:
The maximum height \( h \) can also be expressed in terms of the vertical component of the initial velocity:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
Substituting \( u = 20 \):
\[ h = \frac{20^2 \sin^2 \theta}{2 \cdot 10} \]
\[ h = \frac{400 \sin^2 \theta}{20} \]
\[ h = 20 \sin^2 \theta \quad \text{(Equation 2)} \]

5. **Equating the Two Expressions for Height**:
Now we can set Equation 1 equal to Equation 2:
\[ 20 \cos^2 \theta = 20 \sin^2 \theta \]
Dividing both sides by 20:
\[ \cos^2 \theta = \sin^2 \theta \]

6. **Using the Pythagorean Identity**:
From the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can substitute \( \cos^2 \theta = \sin^2 \theta \):
\[ 2 \sin^2 \theta = 1 \]
\[ \sin^2 \theta = \frac{1}{2} \]
Thus,
\[ \sin \theta = \frac{1}{\sqrt{2}} \quad \text{and} \quad \cos \theta = \frac{1}{\sqrt{2}} \]

7. **Finding Time of Flight**:
The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u \sin \theta}{g} \]
Substituting the values:
\[ T = \frac{2 \cdot 20 \cdot \frac{1}{\sqrt{2}}}{10} \]
\[ T = \frac{40/\sqrt{2}}{10} \]
\[ T = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \, \text{seconds} \]

### Final Answer:
The time of flight is \( 2\sqrt{2} \, \text{seconds} \).