A ball is thrown vertically upwards with a velcotiy of `20 ms^(-1)` from the top of a multi-storey building. The height of the point fromwher the ball is thrown if 25m from the ground. (a) How high the ball will rise ? And (b) how long will it be before the ball hits the ground ? Take. `g=10 ms^(-2)`.

Options

A. `30 m`
B. `20 m`
C. `10 m`
D. `50 m`

Explanation

To solve the problem step by step, we will break it down into two parts: (a) finding how high the ball will rise, and (b) determining how long it will take for the ball to hit the ground.

### Part (a): How high will the ball rise?

1. **Identify the initial conditions**:
- Initial velocity (u) = 20 m/s (upwards)
- Final velocity (v) at the highest point = 0 m/s (the ball stops rising at the peak)
- Acceleration (a) = -10 m/s² (downward due to gravity)

2. **Use the kinematic equation**:
The equation we will use is:
\[ v^2 = u^2 + 2as \]
where \(s\) is the displacement (height gained from the point of release).

3. **Plug in the values**:
\[ 0 = (20)^2 + 2(-10)s \]
This simplifies to:
\[ 0 = 400 - 20s \]

4. **Rearranging the equation**:
\[ 20s = 400 \]
\[ s = \frac{400}{20} = 20 \text{ m} \]

5. **Calculate the total height**:
The total height from the ground is the height of the building plus the height gained:
\[ \text{Total height} = 25 \text{ m} + 20 \text{ m} = 45 \text{ m} \]

### Part (b): How long will it be before the ball hits the ground?

1. **Identify the conditions for the entire motion**:
- Initial velocity (u) = 20 m/s (upwards)
- Displacement (s) = -25 m (the ball falls 25 m to the ground)
- Acceleration (a) = -10 m/s² (downward)

2. **Use the kinematic equation**:
The equation we will use is:
\[ s = ut + \frac{1}{2}at^2 \]

3. **Plug in the values**:
\[ -25 = 20t + \frac{1}{2}(-10)t^2 \]
This simplifies to:
\[ -25 = 20t - 5t^2 \]

4. **Rearranging the equation**:
\[ 5t^2 - 20t - 25 = 0 \]

5. **Dividing the entire equation by 5**:
\[ t^2 - 4t - 5 = 0 \]

6. **Factoring the quadratic equation**:
\[ (t - 5)(t + 1) = 0 \]

7. **Finding the roots**:
This gives us two possible solutions for \(t\):
\[ t = 5 \text{ s} \quad \text{and} \quad t = -1 \text{ s} \]
Since time cannot be negative, we take \(t = 5 \text{ s}\).

### Final Answers:
(a) The maximum height the ball will rise is **45 m**.
(b) The time taken for the ball to hit the ground is **5 seconds**.