https://file.aiquickdraw.com/mathbot/file/1726758066993mas0rotr.jpg

I see two math problems from the image. Let's tackle them one by one.

### Problem 19:
**A car is driving at 80 km/h on a dry summer road when the driver sees an obstacle 60 m ahead. The driver's reaction time is 1.0 seconds. The driver applies the brakes, and the car decelerates at 7.7 m/s².**

**Questions:**
- Does the car stop before hitting the obstacle?
- What is the total distance the car travels from when the driver sees the obstacle until it stops? (Provide the answer in meters, rounded to the nearest whole number.)

#### Solution:
1. **Convert speed to meters per second:**
   \[
   v = 80 \, \text{km/h} = \frac{80 \times 1000}{3600} \, \text{m/s} = 22.22 \, \text{m/s}
   \]

2. **Distance covered during the reaction time:**
   The car travels at constant speed for 1 second during the reaction time.
   \[
   d_{\text{reaction}} = v \times t_{\text{reaction}} = 22.22 \, \text{m/s} \times 1.0 \, \text{s} = 22.22 \, \text{m}
   \]

3. **Braking distance:**
   Using the formula \( v^2 = 2ad \), where \( v \) is the initial velocity, \( a \) is the deceleration, and \( d \) is the distance:
   \[
   d_{\text{braking}} = \frac{v^2}{2a} = \frac{(22.22)^2}{2 \times 7.7} = \frac{493.7}{15.4} \approx 32.06 \, \text{m}
   \]

4. **Total stopping distance:**
   \[
   d_{\text{total}} = d_{\text{reaction}} + d_{\text{braking}} = 22.22 \, \text{m} + 32.06 \, \text{m} = 54.28 \, \text{m}
   \]

Since the total stopping distance (54.28 m) is less than the distance to the obstacle (60 m), **the car stops in time**.

---

### Problem 20:
**You are standing on a 20 m high balcony and want to drop a rose to someone walking toward you at a constant speed of 1.2 m/s. How far must that person be from the point directly below you when the rose is dropped so that the person catches it right when it reaches the ground?**

#### Solution:
1. **Time to fall from 20 m:**
   Using the formula for free fall \( h = \frac{1}{2} g t^2 \), where \( h \) is the height (20 m) and \( g \) is the acceleration due to gravity (9.8 m/s²):
   \[
   20 = \frac{1}{2} \times 9.8 \times t^2 \implies t^2 = \frac{40}{9.8} \approx 4.08 \implies t \approx 2.02 \, \text{seconds}
   \]

2. **Distance the person walks:**
   The person is walking at a constant speed of 1.2 m/s, so in 2.02 seconds, the person walks:
   \[
   d = v \times t = 1.2 \times 2.02 \approx 2.42 \, \text{m}
   \]

Thus, the person must be approximately **2.42 meters** away from the point directly below when the rose is dropped.

---

Do you need more details or have any questions?

### Relative questions:
1. How would the stopping distance change if the car's speed was higher?
2. What if the deceleration was lower? Would the car still stop in time?
3. How does the reaction time influence the total stopping distance in general?
4. What would happen to the falling time if the height was doubled?
5. How would air resistance affect the time for the rose to fall?

### Tip:
Always convert units to be consistent, especially when working with speed and time.

The original questions involve a car's stopping distance and an object being dropped from a height, combined with horizontal motion.

Solve a car stopping distance problem involving a reaction time of 1 second, a speed of 80 km/h, and a deceleration of 7.7 m/s². Also, calculate when to drop a rose from a 20-meter-high balcony for someone walking below at 1.2 m/s. These problems involve kinematics, braking distance, and free fall concepts.

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