Bob drops a ball from rest out of the window of a building. At the very same instant Phyllis launches a ball upwards with an initial speed of 11.0 m/s from a point 20.0 m directly below Bob.
At what height above Phyllis in meters do the balls collide?

Let's solve this problem step by step. We'll need to find the height above Phyllis where the two balls collide.

### Given:
- Bob's ball is dropped from rest, so its initial velocity \( v_{B0} = 0 \) m/s.
- Phyllis's ball is launched upward with an initial velocity \( v_{P0} = 11.0 \) m/s.
- The distance between Bob and Phyllis at the start is 20.0 m.
- Both balls experience acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \), acting downward.

### Step 1: Position equations for the balls

#### Bob's ball
Since Bob's ball is dropped from rest, we can use the following kinematic equation to describe its position after time \( t \):
\[
y_B(t) = h_B - \frac{1}{2} g t^2
\]
where:
- \( y_B(t) \) is the height of Bob's ball above the ground at time \( t \),
- \( h_B \) is Bob's initial height (we’ll set this relative to Phyllis, so \( h_B = 20 \) m),
- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

#### Phyllis's ball
Phyllis's ball is launched upward with an initial velocity. The position of her ball after time \( t \) is given by:
\[
y_P(t) = v_{P0} t - \frac{1}{2} g t^2
\]
where:
- \( y_P(t) \) is the height of Phyllis's ball above the ground at time \( t \),
- \( v_{P0} = 11.0 \, \text{m/s} \) is the initial velocity of Phyllis's ball,
- \( g = 9.8 \, \text{m/s}^2 \).

### Step 2: Condition for collision
At the time \( t \) when the balls collide, their heights must be equal:
\[
y_B(t) = y_P(t)
\]
Substitute the equations for \( y_B(t) \) and \( y_P(t) \):
\[
20 - \frac{1}{2} g t^2 = v_{P0} t - \frac{1}{2} g t^2
\]
Simplify by canceling the \(- \frac{1}{2} g t^2\) terms on both sides:
\[
20 = v_{P0} t
\]
Now, substitute \( v_{P0} = 11.0 \, \text{m/s} \):
\[
20 = 11.0 t
\]
Solving for \( t \):
\[
t = \frac{20}{11} \approx 1.818 \, \text{seconds}
\]

### Step 3: Height above Phyllis at collision
Now, let's find the height at which the collision occurs by plugging \( t = 1.818 \) seconds into Phyllis's position equation:
\[
y_P(t) = v_{P0} t - \frac{1}{2} g t^2
\]
Substitute the values:
\[
y_P(1.818) = 11.0 \times 1.818 - \frac{1}{2} \times 9.8 \times (1.818)^2
\]
\[
y_P(1.818) = 20.0 - \frac{1}{2} \times 9.8 \times 3.305
\]
\[
y_P(1.818) = 20.0 - 16.191 \approx 3.809 \, \text{m}
\]

Thus, the balls collide at a height of approximately **3.81 meters** above Phyllis.

---

Would you like more details or have any questions?

Here are five related questions:
1. What would happen if the initial speed of Phyllis's ball was different?
2. How would the problem change if gravity were stronger or weaker?
3. What if Bob threw his ball downwards instead of dropping it?
4. Can you derive the equation of motion for constant acceleration?
5. What happens to the time of collision if Bob's initial height is doubled?

**Tip**: Always check if forces like air resistance are negligible before applying basic kinematic equations.

Bob drops a ball from rest out of the window of a building. At the very same instant, Phyllis launches a ball upwards with an initial speed of 11.0 m/s from a point 20.0 m directly below Bob. At what height above Phyllis in meters do the balls collide?

Solve a physics problem involving two balls in motion, one dropped from rest and another launched upwards, to find the height above the launch point where they collide. This problem covers key kinematic equations, concepts of projectile motion, and gravity effects. Suitable for students in Grades 9-12.

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