A ball dropped from the top of tower falls first half height of tower in 10 s. The total time spend by ball in air is `["Take "g=10m//s^(2)]`

Options

A. 14.14 s
B. 15.25 s
C. 12.36 s
D. 17.36 s

Answer

To solve the problem step by step, we will follow the physics principles related to motion under gravity.

### Step 1: Understand the Problem
We have a ball dropped from the top of a tower. It falls the first half of the height of the tower in 10 seconds. We need to find the total time the ball spends in the air.

### Step 2: Define Variables
Let:
- \( h \) = total height of the tower
- \( \frac{h}{2} \) = height of the first half of the tower
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t_1 = 10 \, \text{s} \) (time taken to fall the first half)

### Step 3: Use the Equation of Motion
We can use the equation of motion for the first half of the height:
\[ S = ut + \frac{1}{2} a t^2 \]
Where:
- \( S = \frac{h}{2} \)
- \( u = 0 \) (initial velocity, since the ball is dropped)
- \( a = g = 10 \, \text{m/s}^2 \)
- \( t = t_1 = 10 \, \text{s} \)

Substituting the values into the equation:
\[ \frac{h}{2} = 0 \cdot 10 + \frac{1}{2} \cdot 10 \cdot (10)^2 \]
\[ \frac{h}{2} = \frac{1}{2} \cdot 10 \cdot 100 \]
\[ \frac{h}{2} = 500 \, \text{m} \]
Thus, the total height \( h \) is:
\[ h = 1000 \, \text{m} \]

### Step 4: Calculate Total Time to Fall the Entire Height
Now, we need to calculate the total time \( t \) taken to fall the entire height \( h \). We will again use the equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
Where:
- \( S = h = 1000 \, \text{m} \)
- \( u = 0 \)
- \( a = g = 10 \, \text{m/s}^2 \)

Substituting the values:
\[ 1000 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
\[ 1000 = 5t^2 \]
\[ t^2 = \frac{1000}{5} = 200 \]
Taking the square root:
\[ t = \sqrt{200} \approx 14.14 \, \text{s} \]

### Step 5: Conclusion
The total time spent by the ball in the air is approximately \( 14.14 \, \text{s} \).

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