Projectile Motion Worksheet Name: Physics Period: N/A 1. Draw in ... - Free Printable

Let's solve each problem step by step. Since I can't view the image directly, I'll analyze and solve the problems based on the text you've provided.

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Problem 1: Draw velocity components for a cannonball

Instructions:
Draw the x-velocity (\(v_x\)) and y-velocity (\(v_y\)) components for each dot along the path of the cannonball. The first one is done for you.

Explanation:
- In projectile motion, the horizontal component of velocity (\(v_x\)) remains constant (assuming no air resistance).
- The vertical component (\(v_y\)) changes due to gravity: it increases downward as the object falls.
- At the top of the trajectory, \(v_y = 0\), then becomes negative (downward) as it descends.

How to draw:
- For each dot:
- Draw a horizontal arrow (\(v_x\)) of equal length (same magnitude).
- Draw a vertical arrow (\(v_y\)):
- Upward before the peak (positive),
- Zero at the peak,
- Downward after the peak (negative).

> This is a conceptual drawing task. You would sketch these vectors on the diagram.

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Problem 2: Stone thrown horizontally from a cliff

Given:
- Initial horizontal speed: \(v_x = 8.0 \, \text{m/s}\)
- Height of cliff: \(h = 80 \, \text{m}\)
- Acceleration due to gravity: \(g = 9.8 \, \text{m/s}^2\)

Find: How far from the base of the cliff will the stone strike the ground?

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#### Step 1: Find time to fall 80 m (vertical motion)

Use the equation of motion:

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

\[
80 = \frac{1}{2} \times 9.8 \times t^2
\]

\[
t^2 = \frac{80 \times 2}{9.8} = \frac{160}{9.8} \approx 16.327
\]

\[
t \approx \sqrt{16.327} \approx 4.04 \, \text{seconds}
\]

#### Step 2: Use time to find horizontal distance

Horizontal motion has constant velocity:

\[
x = v_x \cdot t = 8.0 \times 4.04 = 32.32 \, \text{m}
\]

✔ Answer: The stone will strike the ground approximately 32.3 meters from the base of the cliff.

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Problem 3: Toy car off a table

Given:
- Height: \(h = 1.25 \, \text{m}\)
- Horizontal distance: \(x = 0.40 \, \text{m}\)
- Initial vertical velocity: \(v_{y0} = 0\) (since it rolls off horizontally)

Find: Time between leaving the table and hitting the floor.

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#### Step 1: Use vertical motion to find time

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

\[
1.25 = \frac{1}{2} \times 9.8 \times t^2
\]

\[
t^2 = \frac{2 \times 1.25}{9.8} = \frac{2.5}{9.8} \approx 0.2551
\]

\[
t \approx \sqrt{0.2551} \approx 0.505 \, \text{seconds}
\]

✔ Answer: The time passed is approximately 0.505 seconds.

> Note: We don’t need the horizontal distance to find time — that’s extra info.

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Problem 4: Bullet fired horizontally at a target

Given:
- Horizontal distance: \(x = 120 \, \text{m}\)
- Initial velocity: \(v_x = 200 \, \text{m/s}\)
- Fired horizontally → initial vertical velocity \(v_{y0} = 0\)
- Both bullet and target are at same height initially

Find:
1. Time to reach the target
2. How far below the target the bullet hits

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#### Part 1: Time to reach target

\[
t = \frac{x}{v_x} = \frac{120}{200} = 0.6 \, \text{seconds}
\]

✔ Time: 0.6 seconds

#### Part 2: Vertical drop during this time

Use vertical motion:

\[
y = \frac{1}{2} g t^2 = \frac{1}{2} \times 9.8 \times (0.6)^2
\]

\[
= 4.9 \times 0.36 = 1.764 \, \text{m}
\]

✔ The bullet hits 1.76 meters below the target.

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✔ Final Answers Summary:

1. Problem 1: Draw constant \(v_x\) and changing \(v_y\) (up → zero → down).
2. Problem 2: Stone lands 32.3 m from base.
3. Problem 3: Time in air is 0.505 s.
4. Problem 4:
- Time to target: 0.6 s
- Drop: 1.76 m below target

Let me know if you'd like diagrams or further explanation!