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

We are given four projectile motion problems. Let’s solve each one step by step.

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Problem 2:
*A stone is thrown horizontally at a speed of 8.0 m/s from the edge of a cliff 80 m in height. How far from the base of the cliff will the stone strike the ground?*

Since it's thrown horizontally, the initial vertical velocity $ v_{y0} = 0 $.
Horizontal velocity $ v_x = 8.0 \, \text{m/s} $ (constant, no air resistance).
Vertical motion is free fall from rest:
Use:
$$
y = v_{y0} t + \frac{1}{2} g t^2
$$
Here, $ y = -80 \, \text{m} $ (downward is negative), $ v_{y0} = 0 $, $ g = -9.8 \, \text{m/s}^2 $ (if we take upward as positive). So:

$$
-80 = 0 \cdot t + \frac{1}{2}(-9.8)t^2 \\
-80 = -4.9 t^2 \\
t^2 = \frac{80}{4.9} \approx 16.3265 \\
t \approx \sqrt{16.3265} \approx 4.04 \, \text{s}
$$

Now horizontal distance:
$$
x = v_x \cdot t = 8.0 \cdot 4.04 = 32.32 \, \text{m}
$$

So answer ≈ 32.3 m

Let me double-check:
$ 4.9 \times (4.04)^2 = 4.9 \times 16.3216 = 79.976 \approx 80 $ — good.

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Problem 3:
*A toy car moves off the edge of a table that is 1.25 m high and lands 0.40 m from the base of the table. How much time passed between the moment the car left the table and the moment it hit the floor?*

Again, horizontal launch → $ v_{y0} = 0 $.
Vertical drop: $ y = -1.25 \, \text{m} $, $ g = -9.8 \, \text{m/s}^2 $

$$
-1.25 = -\frac{1}{2} \cdot 9.8 \cdot t^2 \\
1.25 = 4.9 t^2 \\
t^2 = \frac{1.25}{4.9} \approx 0.2551 \\
t \approx \sqrt{0.2551} \approx 0.505 \, \text{s}
$$

Check: $ 4.9 \times (0.505)^2 = 4.9 \times 0.255 = 1.2495 \approx 1.25 $ — correct.

So time ≈ 0.505 s (or about 0.51 s if rounding to 2 sig figs — but inputs are 1.25 (3 sf) and 0.40 (2 sf), so answer should be 2 sf: 0.51 s)

But note: the horizontal distance (0.40 m) is not needed to find time — only vertical motion determines time of flight for horizontal launch.

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Problem 4:
*A pistol is fired horizontally toward a target 120 m away but at the same height. The bullet’s velocity is 200 m/s. How long does it take the bullet to get to the target? How far below the target does the bullet hit?*

First part: time to reach target (horizontal motion):
$$
t = \frac{d}{v_x} = \frac{120}{200} = 0.6 \, \text{s}
$$

Second part: how far it drops vertically in that time (free fall from rest):
$$
y = \frac{1}{2} g t^2 = \frac{1}{2} \cdot 9.8 \cdot (0.6)^2 = 4.9 \cdot 0.36 = 1.764 \, \text{m}
$$

So bullet hits 1.76 m below target (≈ 1.8 m with 2 sig figs; inputs: 120 m (2 or 3 sf?), 200 m/s (1 or 3 sf?) — safest is 2 sig figs: 1.8 m)

But let’s keep more precision unless instructed otherwise. Since 200 and 120 both have 2 or 3 — 200 could be 2 sf (2.0×10²), 120 could be 2 sf (1.2×10²), so answer: 0.60 s and 1.8 m.

Wait — 120 has 2 or 3? Usually trailing zero without decimal is ambiguous, but in physics problems like this, often treat as 3 sf. However, 200 m/s is likely 3 sf (200. would be explicit). But since problem says “200 m/s” and “120 m”, safest to give answers as:
- Time: 0.600 s
- Drop: 1.76 m

But let’s compute exactly:
$ t = 120 / 200 = 0.6 $ s exactly (since both are whole numbers).
Then drop: $ \frac{1}{2} \cdot 9.8 \cdot 0.36 = 4.9 \cdot 0.36 = 1.764 $ m.

So final: 0.6 s, 1.76 m (or 1.8 m if rounding to 2 sf).

Given typical school expectations, I’ll use 0.60 s and 1.76 m (keeping 3 sig figs where possible).

But the question asks two things:
- How long? → 0.60 s
- How far below? → 1.76 m

Now, back to Problem 1: It says “Draw in the x-velocity and y-velocity components…” — but since this is text-only and user didn’t ask to draw, and the instruction says “Solve the problem accurately”, and only problems 2–4 are numerical, we focus on those.

The user uploaded a worksheet with 4 items, but item 1 is a drawing task — we can’t draw here, and the instruction says “Solve the problem”, implying the calculable ones.

So final answers:

2. 32.3 m
3. 0.505 s (or 0.51 s)
4. Time: 0.60 s, Drop: 1.76 m

Let me verify significant figures:

- Problem 2: 8.0 m/s (2 sf), 80 m (1 or 2 sf? Likely 2 — 8.0 × 10¹), so answer should be 2 sf → 32 m? Wait: 80 has ambiguous sig figs. In many textbooks, “80 m” with no decimal is considered 1 sig fig, but in physics problems like this, they usually intend 2. Given 8.0 has 2, and 80 is likely exact or 2 sf, answer 32 m (2 sf) is safer.

But our calculation gave 32.32 → round to 32 m (2 sf).

However, many teachers accept 32.3 m. Let’s see standard practice: if height is “80 m”, and speed is “8.0 m/s”, the limiting is 8.0 (2 sf) and 80 (maybe 2 sf if written as 8.0×10¹). I think it's fine to give 32 m.

But the problem may expect precise calculation. Since it's a worksheet, likely expects using g = 9.8 and giving 2 decimal places.

Let me recalc with g = 9.8 exactly:

t = √(2h/g) = √(2×80 / 9.8) = √(160/9.8) = √16.32653 = 4.0406 s
x = 8.0 × 4.0406 = 32.3248 → 32.3 m

I’ll go with 32.3 m.

Similarly, problem 3: h = 1.25 m (3 sf), so t = √(2×1.25/9.8) = √(2.5/9.8) = √0.255102 = 0.50508 s → 0.505 s

Problem 4: 120 (3 sf), 200 (3 sf), so t = 0.600 s, drop = 1.764 m → 1.76 m

Final answers:

2. 32.3 m
3. 0.505 s
4. 0.600 s and 1.76 m

Now write Final Answer section with all three (since problem 1 is drawing, not numerical).

But the user said “Solve the problem accurately” — and there are 4 items, but only 2–4 are solvable numerically.

I will provide answers for 2, 3, and 4.

Final Answer:
2. 32.3 m
3. 0.505 s
4. 0.600 s, 1.76 m