A golfer standing on level ground hits a ball with a velocity of `52 m s^-1` at an angle `theta` above the horizontal. If `tan theta = 5//12`, then find the time for which then ball is atleast `15 m` above the ground `(take g = 10 m s^-2)`.

Options

A. 1 s
B. 2 s
C. 3 s
D. 4 s

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the angle theta
Given that \( \tan \theta = \frac{5}{12} \), we can find \( \sin \theta \) and \( \cos \theta \) using the Pythagorean theorem.

Using the sides of the triangle:
- Opposite side (height) = 5
- Adjacent side (base) = 12

Using Pythagoras' theorem:
\[ \text{Hypotenuse} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]

Now, we can find:
\[ \sin \theta = \frac{5}{13}, \quad \cos \theta = \frac{12}{13} \]

### Step 2: Calculate the vertical component of the initial velocity
The initial velocity \( v_0 = 52 \, \text{m/s} \). The vertical component \( v_{y0} \) is given by:
\[ v_{y0} = v_0 \sin \theta = 52 \cdot \frac{5}{13} = 20 \, \text{m/s} \]

### Step 3: Set up the equation of motion
We need to find the time \( t \) for which the ball is at least 15 m above the ground. The equation of motion for vertical displacement is:
\[ y = v_{y0} t - \frac{1}{2} g t^2 \]
where \( g = 10 \, \text{m/s}^2 \) and \( y = 15 \, \text{m} \).

Substituting the values:
\[ 15 = 20t - \frac{1}{2} \cdot 10 t^2 \]
This simplifies to:
\[ 15 = 20t - 5t^2 \]
Rearranging gives us:
\[ 5t^2 - 20t + 15 = 0 \]
Dividing the entire equation by 5:
\[ t^2 - 4t + 3 = 0 \]

### Step 4: Solve the quadratic equation
We can factor the quadratic equation:
\[ (t - 1)(t - 3) = 0 \]
Thus, the solutions for \( t \) are:
\[ t_1 = 1 \, \text{s} \quad \text{and} \quad t_2 = 3 \, \text{s} \]

### Step 5: Determine the time interval
The ball is at least 15 m above the ground between \( t_1 \) and \( t_2 \). Therefore, the time for which the ball is at least 15 m above the ground is:
\[ \Delta t = t_2 - t_1 = 3 \, \text{s} - 1 \, \text{s} = 2 \, \text{s} \]

### Final Answer
The time for which the ball is at least 15 m above the ground is **2 seconds**.

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