A ball is projected form ground with a speed of `20 ms^(-1)` at an angle of 45∘ with horizontal.There is a wall of 25m height at a distance of 10m from the projection point. The ball will hit the wall at a height.

Options

A. 5m
B. 7.5m
C. 10m
D. 12.5m

Explanation

To solve the problem step by step, we will break it down into manageable parts.

### Step 1: Identify the Given Information
- Initial speed of the ball, \( u = 20 \, \text{m/s} \)
- Angle of projection, \( \theta = 45^\circ \)
- Height of the wall, \( H = 25 \, \text{m} \)
- Distance to the wall, \( D = 10 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Resolve the Initial Velocity
The initial velocity can be resolved into horizontal and vertical components:
- Horizontal component, \( u_x = u \cos \theta = 20 \cos 45^\circ = 20 \times \frac{1}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \)
- Vertical component, \( u_y = u \sin \theta = 20 \sin 45^\circ = 20 \times \frac{1}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \)

### Step 3: Calculate the Time to Reach the Wall
Using the horizontal motion equation:
\[ D = u_x \cdot t \]
Substituting the known values:
\[ 10 = 10\sqrt{2} \cdot t \]
Solving for \( t \):
\[ t = \frac{10}{10\sqrt{2}} = \frac{1}{\sqrt{2}} \, \text{s} \]

### Step 4: Calculate the Height of the Ball When It Hits the Wall
Using the vertical motion equation:
\[ y = u_y \cdot t - \frac{1}{2} g t^2 \]
Substituting the known values:
\[ y = (10\sqrt{2}) \cdot \left(\frac{1}{\sqrt{2}}\right) - \frac{1}{2} \cdot 10 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 \]
Calculating each term:
1. \( (10\sqrt{2}) \cdot \left(\frac{1}{\sqrt{2}}\right) = 10 \)
2. \( \frac{1}{2} \cdot 10 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \cdot 10 \cdot \frac{1}{2} = \frac{5}{2} = 2.5 \)

Now, substituting back:
\[ y = 10 - 2.5 = 7.5 \, \text{m} \]

### Conclusion
The ball will hit the wall at a height of **7.5 meters**.

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