A ball is thrown upward with speed 10 m/s from the top to the tower reaches the ground with a speed 20 m/s. The height of the tower is [Take `g=10m//s^(2)`]

Options

A. 10 m
B. 15 m
C. 20 m
D. 25 m

Answer

To solve the problem, we will use the kinematic equation that relates the initial velocity, final velocity, acceleration, and displacement. The equation we will use is:

\[ V^2 = u^2 + 2as \]

Where:
- \( V \) = final velocity (20 m/s)
- \( u \) = initial velocity (10 m/s, but in the downward direction it will be -10 m/s after reaching the maximum height)
- \( a \) = acceleration (which is -g = -10 m/s² since it acts downward)
- \( s \) = displacement (height of the tower, which we will denote as \( h \))

### Step-by-step solution:

1. **Identify the given values:**
- Initial velocity when the ball is thrown upwards, \( u = 10 \, \text{m/s} \)
- Final velocity when the ball hits the ground, \( V = 20 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Determine the initial velocity when the ball is falling:**
- After reaching the maximum height, the ball will start falling downwards with an initial velocity of \( u = -10 \, \text{m/s} \) (downward direction is considered negative).

3. **Use the kinematic equation:**
- We will apply the equation \( V^2 = u^2 + 2as \).
- Here, \( V = 20 \, \text{m/s} \), \( u = -10 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \), and \( s = h \).

4. **Substituting the values into the equation:**
\[ (20)^2 = (-10)^2 + 2(-10)(h) \]
\[ 400 = 100 - 20h \]

5. **Rearranging the equation to solve for \( h \):**
\[ 400 - 100 = -20h \]
\[ 300 = -20h \]
\[ h = \frac{300}{20} = 15 \, \text{m} \]

6. **Conclusion:**
- The height of the tower is \( h = 15 \, \text{m} \).