A ball is thrown horizontally with speed `10 m//s` from the top of tower of height `80 m`. After how much time and at what horizontal distance from the tower it strikes the ground ?

Options

A. `2 s, 40 m`
B. `2 s, 20 m`
C. `4 s, 40 m`
D. `4 s, 80 m`

Explanation

To solve the problem of a ball thrown horizontally from a tower, we can break it down into two parts: the vertical motion and the horizontal motion.

### Step 1: Determine the time of flight
The ball is thrown horizontally from a height of 80 m. We can use the equation of motion for vertical displacement to find the time it takes for the ball to hit the ground.

The equation for vertical motion is:
\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \( s \) is the vertical displacement (80 m)
- \( u \) is the initial vertical velocity (0 m/s, since it is thrown horizontally)
- \( a \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))

Substituting the values into the equation:
\[ 80 = 0 \cdot t + \frac{1}{2} \cdot 9.81 \cdot t^2 \]
\[ 80 = \frac{1}{2} \cdot 9.81 \cdot t^2 \]
\[ 80 = 4.905 \cdot t^2 \]

Now, we can solve for \( t^2 \):
\[ t^2 = \frac{80}{4.905} \]
\[ t^2 \approx 16.3 \]
\[ t \approx \sqrt{16.3} \]
\[ t \approx 4.03 \, \text{seconds} \]

### 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 \( x \) can be calculated using the formula:
\[ x = u_x \cdot t \]

Where:
- \( u_x \) is the horizontal velocity (10 m/s)
- \( t \) is the time of flight (approximately 4.03 seconds)

Substituting the values:
\[ x = 10 \cdot 4.03 \]
\[ x \approx 40.3 \, \text{meters} \]

### Final Answer
The ball strikes the ground after approximately **4.03 seconds** and at a horizontal distance of approximately **40.3 meters** from the tower.

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