A ball is projected horizontally with a velocity of `5ms^(-1)` from the top of a building ` 19.6 m` high. How long will the ball take to hit the ground ?

Explanation

To solve the problem of how long the ball will take to hit the ground when projected horizontally from a height of 19.6 m, we can follow these steps:

### Step 1: Identify the given values
- Height of the building (h) = 19.6 m
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Use the formula for time of flight
The time of flight (t) for an object falling from a height can be calculated using the formula:
\[ t = \sqrt{\frac{2h}{g}} \]

### Step 3: Substitute the values into the formula
Now, we will substitute the values of h and g into the formula:
\[ t = \sqrt{\frac{2 \times 19.6}{9.8}} \]

### Step 4: Simplify the expression
Calculating the numerator:
\[ 2 \times 19.6 = 39.2 \]
Now, substituting this back into the formula:
\[ t = \sqrt{\frac{39.2}{9.8}} \]

### Step 5: Perform the division
Now, we divide:
\[ \frac{39.2}{9.8} = 4 \]

### Step 6: Take the square root
Now, we take the square root of 4:
\[ t = \sqrt{4} = 2 \text{ seconds} \]

### Final Answer
The ball will take **2 seconds** to hit the ground.

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