You playfully throw a coin from the ground at an initial velocity v0 = (8.00i^+ 8.00j^) [m/s] to a wishing well 20.00 [m] away from you when suddenly, a jogger 2.00 [m] tall runs in front of you. As the coin descends, it hits the jogger right on the top of their head. Assuming the coin would have landed in the well, how far is the jogger from the well at this instant?

Let's break down the problem into key components to solve for the distance between the jogger and the well.

### 1. Horizontal Motion Analysis:
We are given the horizontal component of the initial velocity:

\[
v_{0x} = 8.00 \ \text{m/s}
\]

The horizontal distance between you and the well is:

\[
d = 20.00 \ \text{m}
\]

From the equation of motion, the time \( t \) taken to reach the well (in the absence of the jogger) is determined by the constant horizontal velocity since there are no horizontal forces acting on the coin:

\[
x(t) = v_{0x} t
\]
Substituting \( x(t) = 20.00 \ \text{m} \):

\[
20.00 = 8.00 \cdot t
\]

Solving for \( t \):

\[
t = \frac{20.00}{8.00} = 2.50 \ \text{seconds}
\]

This means the coin would have landed in the well in \( 2.50 \) seconds had the jogger not intercepted it.

### 2. Vertical Motion Analysis:
The vertical motion of the coin is influenced by gravity. We are given the vertical component of the initial velocity:

\[
v_{0y} = 8.00 \ \text{m/s}
\]

The vertical position of the coin as a function of time \( y(t) \) can be written as:

\[
y(t) = v_{0y} t - \frac{1}{2} g t^2
\]

where \( g = 9.8 \ \text{m/s}^2 \) is the acceleration due to gravity. We know the jogger is hit on the top of their head, which is 2.00 m above the ground. Therefore, at the time when the coin strikes the jogger, the vertical position of the coin must equal 2.00 m:

\[
2.00 = 8.00 \cdot t - \frac{1}{2} \cdot 9.8 \cdot t^2
\]

This simplifies to:

\[
2.00 = 8.00 t - 4.9 t^2
\]

Rearranging the equation:

\[
4.9 t^2 - 8.00 t + 2.00 = 0
\]

This is a quadratic equation in \( t \). Using the quadratic formula:

\[
t = \frac{-(-8.00) \pm \sqrt{(-8.00)^2 - 4 \cdot 4.9 \cdot 2.00}}{2 \cdot 4.9}
\]

Simplifying:

\[
t = \frac{8.00 \pm \sqrt{64.00 - 39.2}}{9.8} = \frac{8.00 \pm \sqrt{24.8}}{9.8}
\]

\[
t = \frac{8.00 \pm 4.98}{9.8}
\]

This gives two possible solutions for \( t \):

\[
t_1 = \frac{8.00 + 4.98}{9.8} \approx 1.33 \ \text{seconds}
\]
\[
t_2 = \frac{8.00 - 4.98}{9.8} \approx 0.31 \ \text{seconds}
\]

Since the coin is descending when it hits the jogger, we select the larger time:

\[
t = 1.33 \ \text{seconds}
\]

### 3. Distance from Jogger to Well:
Now, we can calculate how far the jogger is from the well at the instant the coin hits them. Using the horizontal motion equation again:

\[
x(t) = v_{0x} \cdot t = 8.00 \cdot 1.33 = 10.64 \ \text{m}
\]

This is the horizontal distance traveled by the coin in 1.33 seconds. Therefore, the distance from the jogger to the well is:

\[
\text{Distance from jogger to well} = 20.00 - 10.64 = 9.36 \ \text{m}
\]

### Final Answer:
The jogger is approximately **9.36 meters** away from the well at the instant the coin hits their head.

---

Do you have any questions or would you like more details? Here are five related questions to deepen your understanding:

1. How would the result change if the jogger were 3 meters tall instead of 2 meters?
2. What would happen if the initial vertical velocity were greater than 8.00 m/s?
3. How do we calculate the maximum height reached by the coin during its trajectory?
4. How would the time change if the horizontal distance to the well were doubled?
5. What effect would air resistance have on the motion of the coin?

**Tip:** In projectile motion, horizontal and vertical components of motion are treated independently, with gravity only affecting vertical motion.

You playfully throw a coin from the ground at an initial velocity v0 = (8.00i^ + 8.00j^) [m/s] to a wishing well 20.00 [m] away from you when suddenly, a jogger 2.00 [m] tall runs in front of you. As the coin descends, it hits the jogger right on the top of their head. Assuming the coin would have landed in the well, how far is the jogger from the well at this instant?

Solve a projectile motion problem involving a coin thrown towards a wishing well. Determine how far a jogger, 2.00 meters tall, is from the well when the coin strikes them. This problem covers horizontal and vertical motion analysis, using kinematics and quadratic equations. Suitable for students in Grades 10-12.

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