A ball is thrown upward from the top of a 25.0 m tall building. The ball's initial speed is 12.0 m/s. At the same instant, a person is running on the ground at a distance of 31.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building ?

Options

A. 8.18 m/s
B. 0.122 m/s
C. 7.20 m/s
D. 0.139 m/s

Answer

To solve the problem, we need to determine the average speed of the person running on the ground so that he can catch the ball at the bottom of the 25.0 m tall building. The ball is thrown upward with an initial speed of 12.0 m/s.

### Step-by-Step Solution:

1. **Identify the variables:**
- Height of the building (h) = 25.0 m
- Initial speed of the ball (u) = 12.0 m/s (upward)
- Acceleration due to gravity (g) = 9.8 m/s² (we can approximate this to 10 m/s² for simplicity)
- Distance of the person from the building (d) = 31.0 m

2. **Determine the time taken for the ball to reach the ground:**
We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, the displacement (s) is -25 m (downward), u = 12 m/s (upward), and a = -g = -10 m/s² (downward).

Plugging in the values:
\[ -25 = 12t - \frac{1}{2} \cdot 10 \cdot t^2 \]
Rearranging gives:
\[ 0 = -5t^2 + 12t + 25 \]
Multiplying through by -1:
\[ 5t^2 - 12t - 25 = 0 \]

3. **Solve the quadratic equation:**
We can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 5 \), \( b = -12 \), and \( c = -25 \).
\[ t = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 5 \cdot (-25)}}{2 \cdot 5} \]
\[ t = \frac{12 \pm \sqrt{144 + 500}}{10} \]
\[ t = \frac{12 \pm \sqrt{644}}{10} \]
\[ t = \frac{12 \pm 25.37}{10} \]
Taking the positive root:
\[ t = \frac{12 + 25.37}{10} = 3.73 \text{ seconds} \]

4. **Calculate the average speed of the person:**
The average speed (v) of the person is given by:
\[ v = \frac{\text{distance}}{\text{time}} \]
Here, the distance is 31 m and the time is 3.73 s:
\[ v = \frac{31}{3.73} \approx 8.31 \text{ m/s} \]

### Final Answer:
The average speed of the person must be approximately **8.31 m/s** to catch the ball at the bottom of the building.

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