Projectile Motion Problems Step by Step

Ignoring air resistance:

• Horizontal (x): no force acts sideways, so horizontal velocity is constant. $a_x = 0$.
• Vertical (y): gravity pulls down, so $a_y = -g = -9.8 \text{ m/s}^2$ (taking up as positive).
• Time is the same in both directions. It links the two halves of the problem.

The equations you need:

$$x = v_x t \qquad v_y = v_{0y} - gt \qquad y = y_0 + v_{0y}t - \tfrac{1}{2}gt^2$$

If the launch is at an angle $\theta$ above horizontal with speed $v_0$, split the velocity first:

$$v_x = v_0 \cos\theta \qquad v_{0y} = v_0 \sin\theta$$

Problem: A ball rolls off the edge of a 19.6 m high cliff with a horizontal speed of 5.0 m/s. How long is it in the air, and how far from the base does it land?

1. List x and y separately.
   • x: $v_x = 5.0$ m/s, $a_x = 0$.
   • y: $v_{0y} = 0$ (launched horizontally), $a_y = -9.8 \text{ m/s}^2$, $\Delta y = -19.6$ m.
2. Use the vertical motion to find time. With $v_{0y} = 0$:

$$-19.6 = -\tfrac{1}{2}(9.8)t^2 \quad\Rightarrow\quad t^2 = \frac{19.6}{4.9} = 4.0 \quad\Rightarrow\quad t = 2.0 \text{ s}$$

3. Use the time in the horizontal motion:

$$x = v_x t = (5.0)(2.0) = 10. \text{ m}$$

4. Bonus, the impact velocity: $v_y = -9.8 \times 2.0 = -19.6$ m/s and $v_x = 5.0$ m/s.

$$v = \sqrt{5.0^2 + 19.6^2} = \sqrt{409.16} = 20.2 \text{ m/s}$$

The angle below horizontal is $\tan^{-1}(19.6 / 5.0) = 75.7^\circ$.

Notice the horizontal speed had nothing to do with the fall time. A ball dropped straight down from the same height would land at the same moment.

Problem: A ball is kicked at 20.0 m/s at $30.0^\circ$ above the horizontal and lands at the same height. Find the time of flight, maximum height, and range.

1. Split the velocity:
   • $v_x = 20.0 \cos 30.0^\circ = 17.32$ m/s
   • $v_{0y} = 20.0 \sin 30.0^\circ = 10.0$ m/s
2. Time to the top. At max height, $v_y = 0$:

$$0 = 10.0 - 9.8t \quad\Rightarrow\quad t_{\text{top}} = 1.02 \text{ s}$$

3. Total time. On level ground the trip down takes as long as the trip up: $t = 2 \times 1.020 = 2.04$ s.
4. Maximum height. Use $v_y^2 = v_{0y}^2 - 2g\Delta y$ with $v_y = 0$:

$$\Delta y = \frac{v_{0y}^2}{2g} = \frac{10.0^2}{2(9.8)} = 5.10 \text{ m}$$

5. Range:

$$x = v_x t = (17.32)(2.041) = 35.3 \text{ m}$$

For level ground you can check with the range formula:

$$R = \frac{v_0^2 \sin 2\theta}{g} = \frac{(20.0)^2 \sin 60.0^\circ}{9.8} = \frac{400 \times 0.8660}{9.8} = 35.3 \text{ m}$$

When the landing height differs from the launch height, solve the vertical equation as a quadratic.

Problem: A ball is thrown from the top of a 10.0 m building at 12.0 m/s, $30.0^\circ$ above the horizontal. How far from the building does it land?

1. Split the velocity: $v_x = 12.0\cos 30.0^\circ = 10.39$ m/s and $v_{0y} = 12.0\sin 30.0^\circ = 6.00$ m/s.
2. Vertical equation. The ball ends 10.0 m below where it started, so $\Delta y = -10.0$ m:

$$-10.0 = 6.00t - 4.9t^2$$

3. Rearrange into standard form: $4.9t^2 - 6.00t - 10.0 = 0$.
4. Quadratic formula:

$$t = \frac{6.00 \pm \sqrt{6.00^2 + 4(4.9)(10.0)}}{2(4.9)} = \frac{6.00 \pm \sqrt{232}}{9.8} = \frac{6.00 \pm 15.23}{9.8}$$

The negative root has no physical meaning, so $t = 21.23 \div 9.8 = 2.17$ s. 5. Horizontal distance: $x = (10.39)(2.166) = 22.5$ m.

The range formula from Example 2 does not apply here, because it assumes the ball lands at the same height it started.

1. Draw the path and choose up as positive.
2. Split the launch velocity into $v_x$ and $v_{0y}$.
3. Make two columns, x and y, and list the known values in each.
4. Find time from whichever direction has enough information (usually y).
5. Use that time in the other direction.
6. Check the result: is the range positive, and is the max height reasonable?

Useful facts at the top of the path: $v_y = 0$, but $v_x$ is unchanged, so the speed there is $v_x$, not zero. Paste it into the VeilGPT solver to check your steps.