A hoodlum throms a stone vertically downward with an initial speed of 15.0 m/s from the roof of a building, 30.0 m above the ground. (a) How long does it take the stone to reach the ground ? (b) What is the speed of the stone at impact ?

Answer

To solve the problem step by step, we will break it down into two parts: (a) finding the time it takes for the stone to reach the ground, and (b) finding the speed of the stone at impact.

### Part (a): Time to Reach the Ground

1. **Identify the known values**:
- Initial speed (u) = -15 m/s (downward)
- Displacement (s) = -30 m (downward)
- Acceleration (a) = -10 m/s² (due to gravity, downward)

2. **Use the kinematic equation**:
We can use the equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
Substituting the known values:
\[ -30 = -15t + \frac{1}{2}(-10)t^2 \]

3. **Simplify the equation**:
\[ -30 = -15t - 5t^2 \]
Rearranging gives:
\[ 5t^2 + 15t - 30 = 0 \]

4. **Divide the entire equation by 5**:
\[ t^2 + 3t - 6 = 0 \]

5. **Use the quadratic formula to solve for t**:
The quadratic formula is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 3\), and \(c = -6\):
\[ t = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \]
\[ t = \frac{-3 \pm \sqrt{9 + 24}}{2} \]
\[ t = \frac{-3 \pm \sqrt{33}}{2} \]

6. **Calculate the positive root**:
Since time cannot be negative, we take the positive root:
\[ t \approx \frac{-3 + 5.744}{2} \approx 1.372 \text{ seconds} \]

### Part (b): Speed of the Stone at Impact

1. **Use the kinematic equation for final velocity**:
We can use the equation:
\[ v = u + at \]
Substituting the known values:
\[ v = -15 + (-10)(1.372) \]
\[ v = -15 - 13.72 \]
\[ v \approx -28.72 \text{ m/s} \]

### Final Answers
- (a) The time taken for the stone to reach the ground is approximately **1.37 seconds**.
- (b) The speed of the stone at impact is approximately **28.72 m/s** (downward).