A cannon ball is projected horizontally from the top of a tower 980 meters high, with a velocity of `50 m//"sec"`. The time of flight is

Options

A. 2 sec
B. `sqrt(2)` sec
C. 1sec
D. `1//2` sec

Answer

To find the time of flight for the cannon ball projected horizontally from the top of a tower, we can use the following steps:

### Step 1: Understand the Problem
The cannon ball is projected horizontally from a height of 980 meters. Since it is projected horizontally, the initial vertical velocity is 0 m/s. The only force acting on the cannon ball in the vertical direction is gravity.

### Step 2: Use the Second Equation of Motion
We can use the second equation of motion for vertical motion:
\[ s = ut + \frac{1}{2}gt^2 \]
where:
- \( s \) is the vertical displacement (height of the tower = 980 m),
- \( u \) is the initial vertical velocity (0 m/s since it is projected horizontally),
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, m/s^2 \)),
- \( t \) is the time of flight.

### Step 3: Substitute the Known Values
Since \( u = 0 \), the equation simplifies to:
\[ s = \frac{1}{2}gt^2 \]
Substituting the known values:
\[ 980 = \frac{1}{2} \times 9.8 \times t^2 \]

### Step 4: Solve for \( t^2 \)
Rearranging the equation gives:
\[ t^2 = \frac{980 \times 2}{9.8} \]
Calculating the right side:
\[ t^2 = \frac{1960}{9.8} \]
\[ t^2 = 200 \]

### Step 5: Calculate \( t \)
Taking the square root of both sides:
\[ t = \sqrt{200} \]
\[ t = 10 \sqrt{2} \, \text{seconds} \]
This can be approximated as:
\[ t \approx 14.14 \, \text{seconds} \]

### Final Answer
The time of flight of the cannon ball is approximately \( 14.14 \) seconds.

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