Two balls are thrown simultaneously from ground with same velocity of `10 ms^(-1)` but different angles of projection with horizontally. Both balls fall at same distance `5sqrt(3)m` from point of projection. What is the time interval between balls striking the ground ?

Options

A. `(sqrt(3)-1)s`
B. `(sqrt(3)+1)s`
C. `sqrt(3)s`
D. 1s

Explanation

To solve the problem of finding the time interval between two balls striking the ground after being thrown simultaneously with the same velocity but at different angles, we can follow these steps:

### Step 1: Understand the given data
- Initial velocity (u) = 10 m/s
- Range (R) = \(5\sqrt{3}\) m
- The angles of projection for the two balls are complementary.

### Step 2: Use the range formula
The range of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \(g\) is the acceleration due to gravity (approximately \(10 \, \text{m/s}^2\)).

### Step 3: Substitute the known values into the range formula
Substituting the values into the range formula:
\[ 5\sqrt{3} = \frac{10^2 \sin(2\theta)}{10} \]
This simplifies to:
\[ 5\sqrt{3} = 10 \sin(2\theta) \]
\[ \sin(2\theta) = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2} \]

### Step 4: Determine the angles of projection
From \(\sin(2\theta) = \frac{\sqrt{3}}{2}\), we find:
\[ 2\theta = 60^\circ \quad \text{or} \quad 2\theta = 120^\circ \]
Thus, the angles of projection are:
\[ \theta_1 = 30^\circ \quad \text{and} \quad \theta_2 = 90^\circ - 30^\circ = 60^\circ \]

### Step 5: Calculate the time of flight for each ball
The time of flight \(T\) for a projectile is given by:
\[ T = \frac{2u \sin(\theta)}{g} \]

#### For the first ball (θ = 30°):
\[ T_1 = \frac{2 \cdot 10 \cdot \sin(30^\circ)}{10} = \frac{20 \cdot \frac{1}{2}}{10} = 1 \, \text{s} \]

#### For the second ball (θ = 60°):
\[ T_2 = \frac{2 \cdot 10 \cdot \sin(60^\circ)}{10} = \frac{20 \cdot \frac{\sqrt{3}}{2}}{10} = \sqrt{3} \, \text{s} \]

### Step 6: Calculate the time interval between the two balls striking the ground
The time interval \(\Delta T\) is given by:
\[ \Delta T = T_2 - T_1 = \sqrt{3} - 1 \, \text{s} \]

### Final Answer
The time interval between the two balls striking the ground is \(\sqrt{3} - 1\) seconds.

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