More Projectile Motion Practice Problems and Solutions | PDF ... - Free Printable

The image you've uploaded contains a projectile motion practice problem set with solutions. Let's go through the problems and their solutions step-by-step, explaining the physics concepts involved.

---

Problem 1: Roadrunner Running Off a Cliff

A roadrunner runs off a cliff with an initial horizontal velocity of 3.5 m/s. The cliff is 300 meters high.

#### a) What are the components of this velocity?

- Horizontal component (Vₓ): Since the roadrunner runs horizontally off the cliff, the initial horizontal velocity is:
$$
V_x = 3.5\ \text{m/s}
$$

- Vertical component (Vᵧ): There is no initial vertical velocity (it doesn't jump up or down), so:
$$
V_y = 0\ \text{m/s}
$$

✔ Answer:
$$
V_x = 3.5\ \text{m/s},\quad V_y = 0\ \text{m/s}
$$

---

#### b) What will be the horizontal velocity 2 seconds after leaving the cliff?

In projectile motion, horizontal velocity remains constant (assuming no air resistance). Gravity only affects vertical motion.

So:
$$
V_x = 3.5\ \text{m/s} \quad \text{(unchanged)}
$$

✔ Answer:
$$
3.5\ \text{m/s} – \text{horizontal velocity is unchanging}
$$

---

#### c) If the cliff is 300 m high, at what time will the roadrunner reach the ground?

We use the vertical motion equation:
$$
h = \frac{1}{2} g t^2
$$
Where:
- $ h = 300\ \text{m} $
- $ g = 10\ \text{m/s}^2 $ (approximated in the solution)
- $ t = ? $

Solve for $ t $:
$$
300 = \frac{1}{2} \cdot 10 \cdot t^2 \\
300 = 5t^2 \\
t^2 = 60 \\
t = \sqrt{60} \approx 7.75\ \text{s}
$$

✔ Answer:
$$
t = 7.75\ \text{seconds}
$$

---

#### d) How far from the cliff will this bird land?

Use horizontal distance formula:
$$
d_x = V_x \cdot t = 3.5\ \text{m/s} \times 7.75\ \text{s} = 27.125\ \text{m}
$$

✔ Answer:
$$
d_x = 27.125\ \text{m}
$$

(Note: The solution says 27.125 m — correct.)

---

#### e) Will the roadrunner land in a pond that starts 25 m away and extends 2.5 m further (i.e., from 25 m to 27.5 m)?

- Landing distance = 27.125 m
- Pond range = 25 m to 27.5 m

Since $ 25 < 27.125 < 27.5 $, yes, it lands in the pond.

✔ Answer:
Yes, the pond is from 25 m to 27.5 m, so the roadrunner will land in the pond.

---

#### f) Final vertical velocity when reaching the ground?

Use:
$$
V_y = V_{y0} + g t = 0 + 10 \cdot 7.75 = 77.5\ \text{m/s}
$$

✔ Answer:
$$
V_y = 77.5\ \text{m/s} \quad \text{(downward)}
$$

---

#### g) Final horizontal velocity?

Horizontal velocity is constant:
$$
V_x = 3.5\ \text{m/s}
$$

✔ Answer:
$$
V_x = 3.5\ \text{m/s}
$$

---

#### h) Total final velocity (magnitude and direction)?

- Magnitude:
$$
V = \sqrt{V_x^2 + V_y^2} = \sqrt{(3.5)^2 + (77.5)^2} = \sqrt{12.25 + 6006.25} = \sqrt{6018.5} \approx 77.579\ \text{m/s}
$$

- Direction (angle below horizontal):
$$
\theta = \tan^{-1}\left(\frac{V_y}{V_x}\right) = \tan^{-1}\left(\frac{77.5}{3.5}\right) \approx \tan^{-1}(22.14) \approx 87.41^\circ
$$

✔ Answer:
- Magnitude: $ V \approx 77.579\ \text{m/s} $
- Direction: $ \theta \approx 87.41^\circ $ below the horizontal

---

Problem 2: Object Dropped from 300 m Height

An object is dropped (not thrown) from 300 m.

#### a) How long does it take to fall?

Same as part (c) above:
$$
h = \frac{1}{2} g t^2 \\
300 = 5t^2 \\
t^2 = 60 \\
t = \sqrt{60} \approx 7.75\ \text{s}
$$

✔ Answer:
$$
t = 7.75\ \text{s}
$$

---

#### b) Compare with Part (c) of Question 1. Why are they the same?

- Both objects fall from 300 m height.
- Both have initial vertical velocity = 0 m/s.
- Vertical motion is governed only by gravity: $ h = \frac{1}{2}gt^2 $
- Horizontal motion does not affect vertical motion.

So, even though one has horizontal velocity and the other doesn’t, the time to fall is identical because vertical motion is independent.

✔ Answer:
They are the same because vertical motions are identical. All objects fall with the same gravitational acceleration, so two objects at the same height with the same initial vertical velocity will reach the ground at the same time.

---

✔ Summary of Key Concepts:

1. Projectile Motion has two independent components:
- Horizontal: Constant velocity (no acceleration if air resistance ignored)
- Vertical: Accelerated motion due to gravity ($ g = 10\ \text{m/s}^2 $ downward)

2. Time of flight depends only on vertical motion and initial vertical velocity.

3. Horizontal distance depends on both horizontal speed and time of flight.

4. Final velocity combines both horizontal and vertical components using vector addition.

---

🔍 Final Notes:

- The solution uses $ g = 10\ \text{m/s}^2 $ for simplicity (common in introductory problems).
- In real life, $ g \approx 9.8\ \text{m/s}^2 $, but here it's fine.
- The answers are correct and well-explained.

Let me know if you'd like a diagram or help visualizing this!