A ball is thrown with a velocity whose horizontal component is `12 ms^-1` from a point `15 m` above the ground and `6 m` away from a verticlewall `18.75 m` high in such a way so as just to clear the wall. At what time will it reach the ground ? `(g = 10 ms^-2)`.

Explanation

To solve the problem, we need to determine the total time taken by the ball to reach the ground after being thrown. We will break the problem down into two parts: the time to reach the wall and the time to fall from the wall to the ground.

### Step 1: Determine the horizontal and vertical components of the initial velocity

Given:
- Horizontal component of velocity, \( u_x = 12 \, \text{m/s} \)
- Height from which the ball is thrown, \( h = 15 \, \text{m} \)
- Distance from the wall, \( d = 6 \, \text{m} \)
- Height of the wall, \( H = 18.75 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

Since the horizontal velocity is given, we can find the time taken to reach the wall using the formula:
\[ t_1 = \frac{d}{u_x} = \frac{6 \, \text{m}}{12 \, \text{m/s}} = 0.5 \, \text{s} \]

### Step 2: Calculate the vertical component of the initial velocity

To just clear the wall, we need to find the vertical component of the velocity when the ball reaches the wall. The vertical displacement when the ball reaches the wall can be calculated using the equation of motion:
\[ H = h - \frac{1}{2} g t_1^2 \]
Substituting the values:
\[ 18.75 = 15 - \frac{1}{2} \cdot 10 \cdot (0.5)^2 \]
Calculating the right side:
\[ 18.75 = 15 - \frac{1}{2} \cdot 10 \cdot 0.25 = 15 - 1.25 = 13.75 \]
This means the ball just clears the wall.

### Step 3: Determine the time taken to fall from the height of the wall to the ground

Now, we need to find the time taken for the ball to fall from the height of the wall (18.75 m) to the ground (0 m). The height it falls is:
\[ h_f = 18.75 \, \text{m} \]
Using the equation of motion for free fall:
\[ h_f = \frac{1}{2} g t_2^2 \]
Rearranging gives:
\[ t_2 = \sqrt{\frac{2h_f}{g}} = \sqrt{\frac{2 \cdot 18.75}{10}} = \sqrt{3.75} \approx 1.936 \, \text{s} \]

### Step 4: Calculate the total time of flight

The total time \( T \) taken by the ball to reach the ground is the sum of the time taken to reach the wall and the time taken to fall from the wall to the ground:
\[ T = t_1 + t_2 = 0.5 \, \text{s} + 1.936 \, \text{s} \approx 2.436 \, \text{s} \]

### Final Answer
The total time taken by the ball to reach the ground is approximately \( 2.436 \, \text{s} \).

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