Projectile Motion interactive worksheet - Free Printable

Let’s solve Problem 1 step by step, as requested.

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Problem 1: Pirate Ship and Cannonball

> A pirate ship, docked at sea a certain distance from the harbor, attacks a fort. The fort retaliates by firing cannonballs at a speed of 82 m/s at an angle of 27° above the horizontal. The cannonballs land on the target pirate ship. How far is the pirate ship from the harbor?

This is a classic projectile motion problem where we are to find:
- Time of flight (tᵣ) — total time the projectile is in the air.
- Range (R) — horizontal distance traveled (this is how far the pirate ship is from the harbor).

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Given:

- Initial velocity: \( v_i = 82 \, \text{m/s} \)
- Launch angle: \( \theta_i = 27^\circ \)
- Acceleration due to gravity: \( g = 9.8 \, \text{m/s}^2 \) *(standard value unless otherwise specified)*

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Formulas Provided:

We’ll use the two formulas given in the worksheet:

1. Time of flight (for symmetric trajectory, same launch and landing height):
\[
t_R = \frac{2 v_i \sin \theta_i}{g}
\]

2. Range (horizontal distance):
\[
R = \frac{v_i^2 \sin(2\theta_i)}{g}
\]

*(Note: \( \sin(2\theta) = 2 \sin\theta \cos\theta \), so this formula comes from combining horizontal velocity and time of flight.)*

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## ✔ Step-by-Step Solution:

Part A: Find \( t_R \)

\[
t_R = \frac{2 \cdot 82 \cdot \sin(27^\circ)}{9.8}
\]

First, calculate \( \sin(27^\circ) \):

\[
\sin(27^\circ) \approx 0.4540
\]

Now plug in:

\[
t_R = \frac{2 \cdot 82 \cdot 0.4540}{9.8} = \frac{74.456}{9.8} \approx 7.60 \, \text{seconds}
\]

✔ So, \( t_R = 7.60 \, \text{s} \)

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Part B: Find \( R \)

\[
R = \frac{(82)^2 \cdot \sin(2 \cdot 27^\circ)}{9.8}
\]

First, compute \( 2\theta_i = 54^\circ \)

\[
\sin(54^\circ) \approx 0.8090
\]

Now compute \( (82)^2 = 6724 \)

\[
R = \frac{6724 \cdot 0.8090}{9.8} = \frac{5439.716}{9.8} \approx 555.07 \, \text{meters}
\]

✔ So, \( R = 555.07 \, \text{m} \)

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## 📝 Final Answers to Fill In:

Given:

- \( v_i = \boxed{82} \, \text{m/s} \)
- \( \theta_i = \boxed{27}^\circ \)
- \( g = \boxed{9.8} \, \text{m/s}^2 \)

Find:

- \( t_R \) and \( R \)

Solution:

Left side (time of flight):

\[
t_R = \frac{2(82 \, \text{m/s}) \sin(27^\circ)}{9.8 \, \text{m/s}^2}
= \frac{2 \cdot 82 \cdot 0.4540}{9.8}
= \frac{74.456}{9.8}
= \boxed{7.60} \, \text{s}
\]

Right side (range):

\[
R = \frac{(82 \, \text{m/s})^2 \sin(54^\circ)}{9.8 \, \text{m/s}^2}
= \frac{6724 \cdot 0.8090}{9.8}
= \frac{5439.716}{9.8}
= \boxed{555.07} \, \text{m}
\]

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## ✔ Final Answer Summary:

- Time of flight: \( \boxed{7.60} \, \text{seconds} \)
- Distance to pirate ship (range): \( \boxed{555.07} \, \text{meters} \)

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> 💡 *Note: If your class uses \( g = 10 \, \text{m/s}^2 \) for simplification, you’d get slightly different values. But since it wasn’t specified, we used the standard \( 9.8 \, \text{m/s}^2 \).*

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## 🔁 Bonus: Check with Horizontal Velocity × Time

Just to verify:

Horizontal velocity: \( v_x = v_i \cos\theta_i = 82 \cdot \cos(27^\circ) \approx 82 \cdot 0.8910 = 73.06 \, \text{m/s} \)

Then, \( R = v_x \cdot t_R = 73.06 \cdot 7.60 \approx 555.26 \, \text{m} \) — matches closely! ✔️

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✔ You can now fill in all the boxes on the worksheet with these values.

Let me know if you want help with Problem 2 next!