Projectile Motion Simulator

The projectile leaves the ground at the origin and lands back at the same height. There is no air resistance, and $g = 9.8$ m/s² downward. The trajectory is drawn in coral on a grid that runs to 95 m horizontally and 48 m vertically, so every setting fits on screen.

The readout lists $v_x$ and $v_{y0}$ (the two components of the launch velocity), the range R, the maximum height H and the flight time, all to two decimal places. On the graph, a teal dot and dashed line mark the peak, which always sits halfway along the range, and a navy dot marks the landing point.

Some comparisons to try: hold the speed fixed and step the angle from 30° up to 60°. The range rises to its largest value at 45° and then falls again, and any two angles that add to 90° land at the same spot. Then hold the angle fixed and double the speed from 10 to 20 m/s: the range and the maximum height both become four times larger, and the flight time doubles.

Split the launch velocity into components: $$v_x = v_0\cos\theta \qquad v_{y0} = v_0\sin\theta$$ The horizontal and vertical motions are independent. Horizontally there is no force, so $v_x$ stays constant and $x = v_x t$. Vertically the acceleration is $-g$, so $v_y = v_{y0} - gt$ and $y = v_{y0}t - \tfrac{1}{2}gt^2$.

At the peak, $v_y = 0$ but $v_x$ is unchanged. Setting $v_y = 0$ gives the time to the top, $v_{y0}/g$, and for a launch that returns to the same height the total flight time is twice that: $$t_{flight} = \frac{2v_{y0}}{g} \qquad H = \frac{v_{y0}^2}{2g} \qquad R = v_x\,t_{flight}$$

Combining these gives $R = \dfrac{v_0^2\sin 2\theta}{g}$, which explains the 45° maximum and the equal ranges for complementary angles. That shortcut is not on the AP Physics 1 equation sheet and only works for level ground, so on the exam build the answer from the kinematics equations, which are on the sheet.

A ball is kicked from level ground at 25 m/s, 40° above the horizontal. Find its range, maximum height and time in the air.

Components. $v_x = 25\cos 40° = 19.15$ m/s and $v_{y0} = 25\sin 40° = 16.07$ m/s.

Time of flight. $t = 2(16.07)/9.8 = 3.28$ s.

Range. $R = (19.15)(3.28) = 62.8$ m.

Maximum height. $H = (16.07)^2 / (2 \times 9.8) = 13.2$ m, reached at $t = 1.64$ s.

Set the sliders to 25 m/s and 40°. The readout should show v_x = 19.15, v_y0 = 16.07, R = 62.81 m, H = 13.18 m and a flight time of 3.28 s. Now move the angle to 50°: the components swap, the range stays at 62.81 m, but H rises to 18.71 m and the flight lasts 3.91 s. The steeper launch spends longer in the air but moves forward more slowly, and the two effects exactly balance.

The ball lands at the same speed it was launched with, 25 m/s, but pointed 40° below the horizontal, because $v_x$ is unchanged and $v_y$ has simply reversed sign.

Projectile motion is part of Topic 1.5, Vectors and Motion in Two Dimensions. Expect questions that compare two launches (which lands first, which goes farther), questions about a horizontally launched object such as a ball rolling off a table, and graph questions where you sketch $v_x$ against time as a flat line and $v_y$ against time as a straight line with slope $-g$.