A ball is projected horizontally from the top of a building 19.6 m high. If the line joining the point of projection to the point where it hits the ground makes an angle of `45^(@)` to the horizontal, the initial velocity of the ball is

Options

A. `4.9 ms^(-1)`
B. `9.1 ms^(-1)`
C. `19.6 ms^(-1)`
D. `14.7 ms^(-1)`

Answer

To solve the problem step by step, we will analyze the motion of the ball projected horizontally from the top of a building.

### Step 1: Understand the Problem
We have a ball projected horizontally from a height of 19.6 meters. The line connecting the point of projection to the point where the ball hits the ground makes a 45-degree angle with the horizontal.

### Step 2: Draw a Diagram
Draw a right triangle where:
- The vertical side represents the height of the building (19.6 m).
- The horizontal side represents the horizontal distance traveled by the ball (range).
- The hypotenuse represents the line from the point of projection to the point of impact, which makes a 45-degree angle with the horizontal.

### Step 3: Find the Time of Flight
The time of flight (T) can be calculated using the formula:
\[ T = \sqrt{\frac{2H}{g}} \]
Where:
- \( H = 19.6 \, \text{m} \) (height of the building)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ T = \sqrt{\frac{2 \times 19.6}{9.8}} = \sqrt{4} = 2 \, \text{s} \]

### Step 4: Relate the Height and Range
Since the angle of projection is 45 degrees, we can use the tangent function:
\[ \tan(45^\circ) = \frac{\text{height}}{\text{range}} = 1 \]
Thus, the height is equal to the range:
\[ \text{Range} = \text{Height} = 19.6 \, \text{m} \]

### Step 5: Calculate the Initial Velocity
The initial horizontal velocity (u) can be found using the formula:
\[ u = \frac{\text{Range}}{T} \]
Substituting the values:
\[ u = \frac{19.6 \, \text{m}}{2 \, \text{s}} = 9.8 \, \text{m/s} \]

### Conclusion
The initial velocity of the ball is \( 9.8 \, \text{m/s} \).

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