A robot is used to perform an experiment. It drops one pebble from an 36m tall building and throws a second pebble down with a 11.4m/s velocity. How much later must the robot throw the 2nd pebble so that they hit the ground at the same time?

We are given two pebbles dropped from the same building, one is dropped freely, and the other is thrown down with an initial velocity. The goal is to find the time delay between the two events so that both pebbles hit the ground at the same time.

### Step 1: Determine the time for the first pebble (dropped freely).

For the first pebble, it is dropped from rest with an initial velocity \( v_0 = 0 \). The height of the building is \( h = 36 \) meters. Using the kinematic equation for free fall:

\[
h = \frac{1}{2} g t_1^2
\]

where:
- \( h = 36 \, \text{m} \),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( t_1 \) is the time it takes for the first pebble to hit the ground.

Solving for \( t_1 \):

\[
36 = \frac{1}{2} \cdot 9.8 \cdot t_1^2
\]

\[
t_1^2 = \frac{36 \times 2}{9.8} = \frac{72}{9.8}
\]

\[
t_1^2 = 7.3469
\]

\[
t_1 = \sqrt{7.3469} \approx 2.71 \, \text{seconds}
\]

### Step 2: Determine the time for the second pebble (thrown with initial velocity).

For the second pebble, it is thrown down with an initial velocity \( v_0 = 11.4 \, \text{m/s} \). Using the kinematic equation:

\[
h = v_0 t_2 + \frac{1}{2} g t_2^2
\]

Substituting the values:

\[
36 = 11.4 \cdot t_2 + \frac{1}{2} \cdot 9.8 \cdot t_2^2
\]

Simplifying:

\[
36 = 11.4 t_2 + 4.9 t_2^2
\]

This is a quadratic equation of the form:

\[
4.9 t_2^2 + 11.4 t_2 - 36 = 0
\]

We can solve this quadratic equation using the quadratic formula:

\[
t_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where:
- \( a = 4.9 \),
- \( b = 11.4 \),
- \( c = -36 \).

Substituting the values:

\[
t_2 = \frac{-11.4 \pm \sqrt{(11.4)^2 - 4 \cdot 4.9 \cdot (-36)}}{2 \cdot 4.9}
\]

\[
t_2 = \frac{-11.4 \pm \sqrt{129.96 + 705.6}}{9.8}
\]

\[
t_2 = \frac{-11.4 \pm \sqrt{835.56}}{9.8}
\]

\[
t_2 = \frac{-11.4 \pm 28.9}{9.8}
\]

Taking the positive root:

\[
t_2 = \frac{-11.4 + 28.9}{9.8} = \frac{17.5}{9.8} \approx 1.79 \, \text{seconds}
\]

### Step 3: Calculate the time delay.

To ensure both pebbles hit the ground at the same time, the second pebble must be thrown \( t_1 - t_2 \) seconds after the first pebble is dropped:

\[
\Delta t = t_1 - t_2 = 2.71 - 1.79 = 0.92 \, \text{seconds}
\]

Thus, the robot must throw the second pebble **0.92 seconds** after dropping the first pebble.

---

Would you like more details on any of the steps, or do you have any questions? Here are some related questions to consider:

1. What if the initial velocity of the second pebble were higher or lower?
2. How would the time delay change if the building were taller?
3. How can you derive the kinematic equations used here?
4. What would happen if the pebbles were thrown upward instead?
5. How would air resistance affect the result?

**Tip:** When dealing with quadratic equations, always check if both roots are physically meaningful based on the context of the problem.

A robot is used to perform an experiment. It drops one pebble from a 36m tall building and throws a second pebble down with an 11.4m/s velocity. How much later must the robot throw the 2nd pebble so that they hit the ground at the same time?

Explore the physics behind dropping one pebble and throwing another from a 36-meter tall building. Learn how to use kinematic equations and quadratic formulas to calculate the time delay, ensuring both pebbles hit the ground at the same time. Suitable for students in Grades 9-11.

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