Heather and Jerry are standing on a bridge 31.4m above a river. Heather throws a rock straight down with a speed of 15.6m/s. Jerry, at exactly the same instant of time, throws a rock straight up with the same speed. Ignore air resistance. How much time elapses between the first splash and the second splash?

We are tasked with finding how much time elapses between the first splash and the second splash when Heather throws a rock downward, and Jerry throws a rock upward, both with the same speed of \( 15.6 \, \text{m/s} \) from a height of \( 31.4 \, \text{m} \) above the river. 

We can break this problem into two parts by analyzing the motion of each rock separately using kinematic equations.

### 1. Rock thrown by Heather (downward):
Since Heather throws the rock downward, the initial velocity of her rock is \( v_{H0} = 15.6 \, \text{m/s} \), and the acceleration is \( g = 9.8 \, \text{m/s}^2 \).

The equation for the position of an object under constant acceleration is:
\[
y = y_0 + v_0 t + \frac{1}{2} a t^2
\]
Where:
- \( y = 0 \) (since the rock hits the water at the river level),
- \( y_0 = 31.4 \, \text{m} \),
- \( v_0 = 15.6 \, \text{m/s} \) (downward initial velocity),
- \( a = 9.8 \, \text{m/s}^2 \).

Plugging in these values, we get:
\[
0 = 31.4 + 15.6 t + \frac{1}{2} (9.8) t^2
\]
Simplifying:
\[
0 = 31.4 + 15.6 t + 4.9 t^2
\]
This is a quadratic equation in \( t \), and we will solve it for the time of splash for Heather's rock.

### 2. Rock thrown by Jerry (upward):
Jerry throws his rock upward with the same speed \( v_{J0} = 15.6 \, \text{m/s} \), but this time the initial velocity is directed upward. The equation for Jerry's rock is the same, except now the initial velocity is positive and the rock must first rise and then fall back down to the river.

For Jerry's rock, the same kinematic equation applies:
\[
0 = 31.4 + 15.6 t - \frac{1}{2} (9.8) t^2
\]
Solving both equations will give us the times for each splash, and the difference between these times will give us the time elapsed between the first and second splash.

Let's solve these quadratic equations.It seems that we encountered a non-real solution while solving the equations. I will refine the process to ensure we only extract the real solutions and ignore any imaginary parts. Let me fix that and try again.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, I can walk you through the process of solving the quadratic equations manually or help with any other questions you might have! 

Would you like to proceed manually? Let me know if you have questions.

Here are 5 relative questions based on this problem:

1. How do you solve quadratic equations in general?
2. How does the motion of an object differ when thrown upward vs. downward?
3. What are the assumptions we make when we ignore air resistance in such problems?
4. How does the height from which an object is dropped affect the time it takes to fall?
5. What is the maximum height reached by Jerry's rock before falling down?

**Tip:** For problems involving kinematic equations, always carefully assign positive and negative signs to velocities and accelerations based on the chosen direction of motion.

Solve the problem where Heather throws a rock down and Jerry throws a rock up, both from a height of 31.4 meters. Learn how to apply kinematic equations and the quadratic formula to find the time difference between the splashes. Suitable for Grades 9-12.

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