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

Options

A. 2s
B. `2sqrt(2)` s
C. `1/2`s
D. `1/sqrt(2)` s

Answer

To solve the problem, we need to find the time of flight of a projectile given that its initial velocity is \( u = 20 \, \text{m/s} \) and the acceleration due to gravity \( g = 10 \, \text{m/s}^2 \). We also know that the potential energy at maximum height is equal to the kinetic energy at that height.

### Step-by-Step Solution:

1. **Understanding the Problem**:
At maximum height, the vertical component of the velocity becomes zero. The potential energy (PE) at maximum height is equal to the kinetic energy (KE) at that height.

2. **Using the Energy Conservation Principle**:
The potential energy at maximum height can be expressed as:
\[ PE = mgh \]
where \( h \) is the maximum height, \( m \) is the mass of the projectile, and \( g \) is the acceleration due to gravity.

The kinetic energy at maximum height is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \( v \) is the horizontal component of the velocity at maximum height.

3. **Finding the Maximum Height**:
The maximum height \( h \) can be expressed in terms of the initial velocity \( u \) and the angle of projection \( \theta \):
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]

4. **Finding the Horizontal Velocity**:
The horizontal component of the velocity at maximum height is:
\[ v = u \cos \theta \]

5. **Setting Potential Energy Equal to Kinetic Energy**:
According to the problem:
\[ mgh = \frac{1}{2} mv^2 \]
Substituting the expressions for \( h \) and \( v \):
\[ mg \left(\frac{u^2 \sin^2 \theta}{2g}\right) = \frac{1}{2} m (u \cos \theta)^2 \]
Simplifying this gives:
\[ \frac{u^2 \sin^2 \theta}{2} = \frac{1}{2} u^2 \cos^2 \theta \]
Cancelling \( \frac{1}{2} u^2 \) from both sides (assuming \( u \neq 0 \)):
\[ \sin^2 \theta = \cos^2 \theta \]
This implies:
\[ \tan^2 \theta = 1 \quad \Rightarrow \quad \theta = 45^\circ \]

6. **Calculating Time of Flight**:
The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u \sin \theta}{g} \]
Substituting \( u = 20 \, \text{m/s} \), \( \theta = 45^\circ \), and \( g = 10 \, \text{m/s}^2 \):
\[ T = \frac{2 \times 20 \times \sin 45^\circ}{10} \]
Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} \):
\[ T = \frac{40 \times \frac{1}{\sqrt{2}}}{10} = \frac{4 \sqrt{2}}{2} = 2\sqrt{2} \, \text{seconds} \]

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

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