A ball is thrown horizontally with a speed of `20 m//s` from the top of a tower of height 100m
Horizontal distance travelled by the before it strickes the ground is

Options

A. `10sqrt5m`
B. `20 sqrt5m`
C. `30 sqrt5m`
D. `40 sqrt5m`

Explanation

To solve the problem of finding the horizontal distance traveled by a ball thrown horizontally from the top of a tower, we can break the problem into two parts: the vertical motion and the horizontal motion.

### Step 1: Determine the time of flight
The first step is to calculate how long it takes for the ball to fall from the height of the tower to the ground. We can use the second equation of motion for vertical displacement:

\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \(s\) is the vertical displacement (100 m),
- \(u\) is the initial vertical velocity (0 m/s, since the ball is thrown horizontally),
- \(a\) is the acceleration due to gravity (approximately \(10 \, \text{m/s}^2\)),
- \(t\) is the time in seconds.

Substituting the values into the equation:

\[ 100 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]

This simplifies to:

\[ 100 = 5 t^2 \]

Now, solving for \(t^2\):

\[ t^2 = \frac{100}{5} = 20 \]

Taking the square root gives:

\[ t = \sqrt{20} \approx 4.47 \, \text{s} \]

### Step 2: Calculate the horizontal distance
Now that we have the time of flight, we can calculate the horizontal distance traveled by the ball. The horizontal distance can be calculated using the formula:

\[ \text{Horizontal distance} = \text{horizontal velocity} \times \text{time} \]

Given that the horizontal velocity \(v_x = 20 \, \text{m/s}\) and using the time we just calculated:

\[ \text{Horizontal distance} = 20 \, \text{m/s} \times \sqrt{20} \, \text{s} \]

Substituting the value of \(\sqrt{20}\):

\[ \text{Horizontal distance} = 20 \times 4.47 \approx 89.4 \, \text{m} \]

### Final Answer
The horizontal distance traveled by the ball before it strikes the ground is approximately **89.4 meters**.

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