From the top of a tower 20m 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. `10ms^(-1)`
B. `4ms^(-1)`
C. `15ms^(-1)`
D. `3ms^(-1)`

Explanation

To solve the problem, we need to analyze the motion of the ball thrown horizontally from the top of a tower. The tower is 20 meters high, and we know that the line joining the point of projection to the point where it hits the ground makes an angle of 45 degrees with the horizontal.

### Step 1: Understand the Geometry of the Problem
When the ball is thrown horizontally from a height of 20 meters, it will fall under the influence of gravity while moving horizontally. The trajectory of the ball will form a right triangle where:
- The vertical side (height of the tower) is 20 m.
- The horizontal side is the horizontal distance (range) traveled by the ball before it hits the ground.
- The hypotenuse is the line joining the point of projection to the point where it hits the ground, which makes a 45-degree angle with the horizontal.

### Step 2: Use the Properties of the Triangle
Since the angle is 45 degrees, the horizontal distance (let's denote it as \( R \)) and the height (20 m) must be equal because in a 45-degree right triangle, the two legs are equal. Thus:
\[ R = 20 \, \text{m} \]

### Step 3: Calculate the Time of Flight
The time \( t \) it takes for the ball to fall 20 m can be calculated using the formula for free fall:
\[ h = \frac{1}{2} g t^2 \]
where \( h = 20 \, \text{m} \) and \( g = 9.81 \, \text{m/s}^2 \).

Rearranging the formula to solve for \( t \):
\[ 20 = \frac{1}{2} \times 9.81 \times t^2 \]
\[ t^2 = \frac{20 \times 2}{9.81} \]
\[ t^2 = \frac{40}{9.81} \]
\[ t = \sqrt{\frac{40}{9.81}} \]
\[ t \approx 2.02 \, \text{s} \]

### Step 4: Calculate the Initial Velocity
The horizontal distance \( R \) traveled by the ball can also be expressed in terms of the initial horizontal velocity \( u \) and the time of flight \( t \):
\[ R = u \cdot t \]
Substituting the values we have:
\[ 20 = u \cdot 2.02 \]
\[ u = \frac{20}{2.02} \]
\[ u \approx 9.90 \, \text{m/s} \]

### Final Answer
The initial velocity of the ball is approximately \( 9.90 \, \text{m/s} \).

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