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 m//s`
B. `4.9m//s`
C. `14.7m//s`
D. `2.8m//s `

Answer

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

### Step 1: Understanding the Problem
We have a ball thrown horizontally from the top of a tower that is 19.6 m high. The ball hits the ground at an angle of 45° with the horizontal. We need to find the initial horizontal velocity of the ball.

### Step 2: Analyzing the Geometry
Since the angle made with the horizontal is 45°, the horizontal distance (x) covered by the ball when it hits the ground is equal to the vertical height (h) from which it was thrown. Thus, we have:
- Height of the tower (h) = 19.6 m
- Horizontal distance (x) = 19.6 m

### Step 3: Using the Equation of Motion for Vertical Motion
For vertical motion, we can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s \) = vertical displacement = 19.6 m
- \( u \) = initial vertical velocity = 0 m/s (since it is thrown horizontally)
- \( a \) = acceleration due to gravity = 9.8 m/s²
- \( t \) = time of flight (which we need to find)

Substituting the values into the equation:
\[ 19.6 = 0 + \frac{1}{2} \times 9.8 \times t^2 \]
This simplifies to:
\[ 19.6 = 4.9 t^2 \]

### Step 4: Solving for Time (t)
Now we can solve for \( t^2 \):
\[ t^2 = \frac{19.6}{4.9} \]
\[ t^2 = 4 \]
Taking the square root of both sides:
\[ t = 2 \text{ seconds} \]

### Step 5: Finding the Horizontal Velocity (u_x)
Now, we can find the horizontal velocity using the formula:
\[ v = \frac{x}{t} \]
Where:
- \( x \) = horizontal distance = 19.6 m
- \( t \) = time of flight = 2 s

Substituting the values:
\[ u_x = \frac{19.6}{2} = 9.8 \text{ m/s} \]

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

### Final Answer
The initial velocity of the ball is **9.8 m/s**.

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