AP Physics C: Mechanics FRQ Practice

⚙️AP Physics C: Mechanics

Practice FRQ 1 of 81/8

1. A ball is launched from ground level with initial speed v0 at various launch angles above the horizontal on level ground. Air resistance is negligible. The coordinate system has its origin at the launch point, with the +x-direction horizontal and the +y-direction vertically upward, as shown in Figure 1.

Figure 1. Projectile launched from the origin with initial speed v₀ at angle θ above horizontal; +x along level ground, +y upward; dashed parabolic trajectory returns to y = 0.

Single-panel physics diagram (no graph tick marks).

Overall layout:
- The bottom of the panel contains a straight, perfectly horizontal ground line spanning the full width of the panel. The ground line is the x-axis reference (y = 0).
- The coordinate axes originate at a point on the ground line located one-quarter of the panel width from the left edge. This point is the launch point and the origin.

Axes:
- From the origin, draw the +x-axis as a horizontal ray coincident with the ground line, pointing to the right. At the far right end of this ray, place an arrowhead and the visible text label “+x”.
- From the origin, draw the +y-axis as a vertical ray pointing upward. At the top end of this vertical ray, place an arrowhead and the visible text label “+y”.
- Place a small solid dot exactly at the axes intersection to mark the origin, and place the visible text label “Origin (launch point)” slightly above and left of the dot with a thin leader line to the dot.

Projectile at launch:
- Draw a small filled circle (the ball) centered exactly on the origin dot.

Initial velocity vector:
- From the center of the ball at the origin, draw a straight arrow representing the initial velocity.
- The arrow must point into the first quadrant (up and to the right) and must form an acute angle θ measured counterclockwise from the +x-axis.
- The arrowhead is at the upper-right end; the tail is exactly at the origin.
- Place the visible text label “v₀” along the arrow shaft near its middle.

Angle marker:
- Draw a circular arc centered at the origin, starting along the +x direction and ending along the initial velocity direction.
- Place the visible text label “θ” centered just outside this arc.

Trajectory:
- Draw a single dashed parabolic arc representing the projectile’s path.
- The arc must start at the origin on the ground line, rise upward to a single highest point (apex) located to the right of the origin, and then descend back to intersect the ground line again at a second point to the right of the origin.
- The parabola is concave downward everywhere.
- The dashed trajectory must be drawn so that its initial tangent direction at the origin matches the direction of the initial velocity arrow.
- The landing point is unlabeled but clearly shown as the second intersection of the dashed curve with the ground line.

Styling:
- Axes and vectors are solid black lines; trajectory is a black dashed line.
- No numerical values are printed in this figure (only variable labels: +x, +y, v₀, θ, and “Origin (launch point)”).

Figure 2. Velocity vectors at launch and at the highest point for θ = 30°, 45°, and 60°. All launch speeds have equal magnitude v₀; at the highest point v_y = 0.

Two-row by three-column array of identical square vector grids, with uniform spacing between grids. Each grid contains perpendicular axes crossing at the exact center, forming a plus sign.

Shared grid formatting (applies to all 6 grids):
- Each grid is a square frame.
- Inside each frame, draw a horizontal axis and a vertical axis that intersect at the exact center of the square.
- At the center intersection, place a small open circle (the “zero-velocity point”). This is the mandatory tail location for all arrows.
- Place a visible “0” label adjacent to the center marker (slightly below-left of the intersection) to denote zero velocity.
- Provide tick marks on each axis at equal spacing: two tick marks on the positive side and two on the negative side for both x and y directions. Do not label these ticks with numbers (only the center is labeled “0”).
- Put small arrowheads at the positive ends of the horizontal and vertical axes (right end for +x-velocity direction, top end for +y-velocity direction). Do not write “x” or “y” on the grids.

Panel titles (visible text above each grid):
Top row, left to right:
1) “At launch (θ = 30°)”
2) “At launch (θ = 45°)”
3) “At launch (θ = 60°)”
Bottom row, left to right:
4) “At highest point (θ = 30°)”
5) “At highest point (θ = 45°)”
6) “At highest point (θ = 60°)”

Vector drawing rules (must be enforced exactly):
- Every arrow starts at the central zero-velocity point.
- Arrow length represents speed magnitude.
- All three launch arrows must have EXACTLY equal length to each other (same v₀).
- At the highest point, vertical component is zero, so each highest-point arrow is purely horizontal to the right, with length proportional to the horizontal component v₀ cos(θ).
- No arrow should point leftward; all horizontal components are to the right.

Top row vectors (launch):
1) At launch (θ = 30°): (already provided in the prompt, but specify precisely)
- Draw an arrow of length L (define L as the common launch-arrow length) from the center, pointing up-right at exactly 30° above the positive horizontal axis.
- The arrowhead is on the 30° ray; tail at center.

2) At launch (θ = 45°):
- Draw an arrow from the center with EXACT length L (identical to the 30° launch arrow length).
- Direction: exactly 45° above the positive horizontal axis (diagonal at equal rise and run).

3) At launch (θ = 60°):
- Draw an arrow from the center with EXACT length L.
- Direction: exactly 60° above the positive horizontal axis (steeper than 45°, closer to vertical but still clearly rightward).

Bottom row vectors (highest point):
4) At highest point (θ = 30°):
- Draw a purely horizontal arrow to the right from the center (0° above horizontal).
- Length must be (cos 30°) times L. This arrow must be longer than the highest-point arrows for 45° and 60°.

5) At highest point (θ = 45°):
- Draw a purely horizontal arrow to the right from the center.
- Length must be (cos 45°) times L.
- This arrow must be shorter than the θ = 30° highest-point arrow and longer than the θ = 60° highest-point arrow.

6) At highest point (θ = 60°):
- Draw a purely horizontal arrow to the right from the center.
- Length must be (cos 60°) times L, i.e., exactly one-half of the launch-arrow length L.
- This is the shortest of the three highest-point arrows.

Zero-length rule:
- Not used in this figure because none of the velocities are zero at the highest point (all have nonzero horizontal components). Therefore, do not place a dot-only marker in any panel.

Styling:
- All velocity arrows are solid black with identical line thickness and identical arrowhead style.
- Grids and axes are thin black lines; titles are centered above each grid.

i. The diagrams in Figure 2 can be used to represent the velocity vectors of the ball at launch and at the highest point of its trajectory for three different launch angles: θ = 30°, θ = 45°, and θ = 60°. The velocity vector at launch for θ = 30° is shown.

Draw arrows on the remaining grids to represent the velocity vectors at launch for θ = 45° and θ = 60°, and at the highest point of the trajectory for all three launch angles.

• Arrows should start at the zero-velocity point at the center of each grid.

• The length of the arrows should be proportional to the relative magnitudes of the velocities.

• All launch velocities have the same magnitude v0.

• Represent an arrow of zero length by drawing a dot at zero.

ii. For a launch angle θ = 60°, the ball is in the air for a total time tf. During the time interval from launch to the highest point, the vertical position of the ball as a function of time is modeled by

y(t) = v_0 \sin(60°)t - \frac{1}{2}gt^2

, where g is the acceleration due to gravity.

Derive an expression for the maximum height H reached by the ball. Express your answer in terms of v0, g, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

B. Derive an expression for tan(α) in terms of physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information. Consider a new scenario where the ball is launched with the same initial speed v0 but at a new launch angle α. In this scenario, the horizontal distance traveled by the ball (the range) is observed to be three times the maximum height reached.