HW 18 Mixed Projectile Motion Worksheet | PDF - Free Printable

Let’s solve each problem step by step, using the principles of projectile motion. In projectile motion, horizontal and vertical motions are independent. We assume air resistance is negligible and gravity \( g = 9.8 \, \text{m/s}^2 \) (unless otherwise specified).

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## Problem 1:
> A stone is thrown horizontally from a cliff 30 m high with an initial speed of 20 m/s. How far from the cliff does the stone strike the ground?

Step 1: Analyze vertical motion to find time in air

Since it's thrown horizontally, the initial vertical velocity \( v_{y0} = 0 \).

Use the equation for vertical displacement:
\[
y = v_{y0} t + \frac{1}{2} g t^2
\]
Here, \( y = 30 \, \text{m} \) (downward, so we take it as positive), \( v_{y0} = 0 \), \( g = 9.8 \, \text{m/s}^2 \)

\[
30 = 0 + \frac{1}{2} (9.8) t^2
\Rightarrow 30 = 4.9 t^2
\Rightarrow t^2 = \frac{30}{4.9} \approx 6.1224
\Rightarrow t \approx \sqrt{6.1224} \approx 2.474 \, \text{seconds}
\]

Step 2: Use horizontal motion to find distance

Horizontal velocity is constant: \( v_x = 20 \, \text{m/s} \)

\[
x = v_x \cdot t = 20 \cdot 2.474 \approx 49.48 \, \text{m}
\]

✔ Answer: The stone strikes the ground approximately 49.5 meters from the cliff.

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## Problem 2:
> A ball rolls off the edge of a table 1.44 m above the floor and strikes the floor at a point 2 m horizontally from the edge of the table.

Part A: What is the time the ball was in the air?

Again, vertical motion determines time. Initial vertical velocity = 0.

\[
y = \frac{1}{2} g t^2
\Rightarrow 1.44 = \frac{1}{2} (9.8) t^2
\Rightarrow 1.44 = 4.9 t^2
\Rightarrow t^2 = \frac{1.44}{4.9} \approx 0.2939
\Rightarrow t \approx \sqrt{0.2939} \approx 0.542 \, \text{seconds}
\]

✔ Answer A: The ball was in the air for approximately 0.54 seconds.

Part B: What is the initial velocity of the ball?

Initial velocity is purely horizontal since it rolled off the table.

\[
v_x = \frac{\text{horizontal distance}}{\text{time}} = \frac{2 \, \text{m}}{0.542 \, \text{s}} \approx 3.69 \, \text{m/s}
\]

✔ Answer B: The initial velocity of the ball is approximately 3.69 m/s.

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## Problem 3:
> A golfer hits a ball and gives it an initial velocity of 40 m/s, at an angle of 30° above the horizontal.

Part A: How long does the ball stay in the air?

We need to find the total time of flight for a projectile launched and landing at the same height.

The formula for time of flight is:
\[
T = \frac{2 v_0 \sin\theta}{g}
\]

Where:
- \( v_0 = 40 \, \text{m/s} \)
- \( \theta = 30^\circ \)
- \( g = 9.8 \, \text{m/s}^2 \)

\[
T = \frac{2 \cdot 40 \cdot \sin(30^\circ)}{9.8}
= \frac{80 \cdot 0.5}{9.8}
= \frac{40}{9.8} \approx 4.082 \, \text{seconds}
\]

✔ Answer A: The ball stays in the air for approximately 4.08 seconds.

Part B: How far horizontally (the range) does the ball travel before hitting the ground?

Range formula:
\[
R = \frac{v_0^2 \sin(2\theta)}{g}
\]

Or, alternatively: \( R = v_x \cdot T \), where \( v_x = v_0 \cos\theta \)

Let’s use the second method:

\[
v_x = 40 \cdot \cos(30^\circ) = 40 \cdot \frac{\sqrt{3}}{2} \approx 40 \cdot 0.8660 = 34.64 \, \text{m/s}
\]

\[
R = v_x \cdot T = 34.64 \cdot 4.082 \approx 141.4 \, \text{m}
\]

Using the range formula to verify:

\[
R = \frac{(40)^2 \cdot \sin(60^\circ)}{9.8}
= \frac{1600 \cdot \frac{\sqrt{3}}{2}}{9.8}
= \frac{1600 \cdot 0.8660}{9.8}
\approx \frac{1385.6}{9.8} \approx 141.4 \, \text{m}
\]

✔ Answer B: The ball travels approximately 141.4 meters horizontally.

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## ✔ Final Answers Summary:

1. Stone strikes ground ≈ 49.5 m from cliff.

2A. Ball in air for ≈ 0.54 s.
2B. Initial velocity ≈ 3.69 m/s.

3A. Time in air ≈ 4.08 s.
3B. Horizontal range ≈ 141.4 m.

Let me know if you’d like diagrams or derivations of the formulas!