A ball is projected obliquely with a velocity 49 `ms^(-1)` strikes the ground at a distance of 245 m from the point of projection. It remained in air for

Options

A. 10 sec
B. `5sqrt(2)` sec
C. 3 sec
D. 2.5 sec

Answer

To solve the problem step by step, we will follow the physics of projectile motion.

### Step 1: Understand the Problem
We have a ball projected obliquely with an initial velocity of 49 m/s, and it strikes the ground at a horizontal distance (range) of 245 m. We need to find the time of flight of the ball.

### Step 2: Identify Relevant Equations
For projectile motion, we can use the following equations:
1. **Range Formula**:
\[ R = \frac{v^2 \sin 2\theta}{g} \]
where \( R \) is the range, \( v \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

2. **Time of Flight Formula**:
\[ T = \frac{R}{v \cos \theta} \]

### Step 3: Calculate the Angle of Projection
From the range formula, we can rearrange it to find \( \sin 2\theta \):
\[ R = \frac{v^2 \sin 2\theta}{g} \implies \sin 2\theta = \frac{R \cdot g}{v^2} \]
Substituting the known values:
- \( R = 245 \, \text{m} \)
- \( v = 49 \, \text{m/s} \)
- \( g = 9.8 \, \text{m/s}^2 \)

Calculating \( \sin 2\theta \):
\[ \sin 2\theta = \frac{245 \cdot 9.8}{49^2} \]
Calculating \( 49^2 = 2401 \):
\[ \sin 2\theta = \frac{2401}{2401} = 1 \]
Since \( \sin 2\theta = 1 \), we have:
\[ 2\theta = 90^\circ \implies \theta = 45^\circ \]

### Step 4: Calculate Time of Flight
Now we can use the time of flight formula:
\[ T = \frac{R}{v \cos \theta} \]
Substituting the values:
- \( R = 245 \, \text{m} \)
- \( v = 49 \, \text{m/s} \)
- \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)

Calculating \( T \):
\[ T = \frac{245}{49 \cdot \frac{1}{\sqrt{2}}} = \frac{245 \sqrt{2}}{49} \]
Calculating \( \frac{245}{49} = 5 \):
\[ T = 5 \sqrt{2} \, \text{seconds} \]

### Final Answer
The ball remained in the air for \( 5\sqrt{2} \) seconds.

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