From the the top of a tower 19.6 m high, a ball is thrown horizontally. If the line joining the point of projection to the point where it hits the ground makes an angle of `45^(@)` with the horizontal, then the initial velocity of the ball is:-

Options

A. `9.8 ms^(-1)`
B. `4.9ms^(-1)`
C. `14.7ms^(-1)`
D. `2.8ms^(-1)`

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Problem
We have a tower of height \( h = 19.6 \, \text{m} \) from which a ball is thrown horizontally. The line joining the point of projection to the point where it hits the ground makes an angle of \( 45^\circ \) with the horizontal.

### Step 2: Draw a Diagram
Draw a right triangle where:
- The vertical side represents the height of the tower (19.6 m).
- The horizontal side represents the horizontal distance traveled by the ball before it hits the ground.
- The hypotenuse represents the line from the point of projection to the point of impact, making a \( 45^\circ \) angle with the horizontal.

### Step 3: Use the Properties of the Triangle
Since the angle is \( 45^\circ \), the horizontal distance \( x \) traveled by the ball is equal to the height \( h \):
\[ x = h = 19.6 \, \text{m} \]

### Step 4: Calculate the Time of Flight
The time of flight \( t \) can be calculated using the formula for free fall:
\[ h = \frac{1}{2} g t^2 \]
Rearranging gives:
\[ t = \sqrt{\frac{2h}{g}} \]
Substituting \( h = 19.6 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \):
\[ t = \sqrt{\frac{2 \times 19.6}{9.8}} = \sqrt{4} = 2 \, \text{s} \]

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

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

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