A golfer standing on the ground hits a ball with a velocity of `52 m//s` at an angle `theta` above the horizontal if `tan theta=5/12` find the time for which the ball is at least `15m` above the ground?
`(g=10m//s^(2))`

Explanation

To solve the problem, we need to find the time for which the ball is at least 15 meters above the ground after being hit by the golfer. We will follow these steps:

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

- Using the triangle formed by the opposite side (5) and adjacent side (12):
- Hypotenuse \( h = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \)
- Therefore, \( \sin \theta = \frac{5}{13} \) and \( \cos \theta = \frac{12}{13} \).

### Step 2: Calculate the initial vertical velocity component (u_y)
The initial velocity \( u \) is given as 52 m/s. The vertical component of the initial velocity \( u_y \) can be calculated as:
\[ u_y = u \sin \theta = 52 \times \frac{5}{13} = 20 \, \text{m/s} \]

### Step 3: Write the equation for vertical motion
The height \( S_y \) of the ball at time \( t \) can be expressed using the equation of motion:
\[ S_y = u_y t - \frac{1}{2} g t^2 \]
Substituting \( S_y = 15 \, \text{m} \), \( u_y = 20 \, \text{m/s} \), and \( g = 10 \, \text{m/s}^2 \):
\[ 15 = 20t - \frac{1}{2} \times 10 t^2 \]
This simplifies to:
\[ 15 = 20t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 20t + 15 = 0 \]
Dividing through by 5:
\[ t^2 - 4t + 3 = 0 \]

### Step 4: Solve the quadratic equation
To find the values of \( t \), we can factor the quadratic:
\[ (t - 1)(t - 3) = 0 \]
Thus, the solutions are:
\[ t = 1 \, \text{s} \quad \text{and} \quad t = 3 \, \text{s} \]

### Step 5: Determine the time interval above 15 meters
The ball is at a height of 15 meters at \( t = 1 \, \text{s} \) and \( t = 3 \, \text{s} \). The ball will be above 15 meters between these two times.

Thus, the time for which the ball is at least 15 meters above the ground is:
\[ t = 3 \, \text{s} - 1 \, \text{s} = 2 \, \text{s} \]

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

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