Answer at a Glance
For a projectile launched at speed v and angle θ, the height at which its speed equals 0.5v is h = (3 v^2 sin^2 θ) / (8 g). This is derived by equating the speed at height h to 0.5v and using conservation of energy or kinematic equations. The height depends on the launch angle and initial speed; at θ = 90°, h = 3 v^2 / (8 g). For typical classroom values (e.g., v = 20 m/s), h is roughly a few meters, well within the upward and downward portions of the trajectory.
What This Question Is Asking
The question asks: at what height h above the ground does a projectile have a speed of 0.5v? Here, v is the initial launch speed, and the projectile moves under constant gravity g (≈ 9.81 m/s²) near Earth’s surface, neglecting air resistance. The target speed is one half of the launch speed. Because speed is the magnitude of velocity, this involves both vertical and horizontal components. The solution links initial conditions, kinematics, and energy principles to find the specific height(s) where this condition holds.
Key Concepts and Definitions
- Projectile motion: Motion of an object launched into the air, influenced only by gravity (idealized without air resistance).
- Speed vs. velocity: Speed is the magnitude of velocity; horizontal and vertical components combine vectorially to determine instantaneous speed.
- Initial speed v: The speed at launch, determined by initial velocity components v_x = v cos θ and v_y = v sin θ.
- Height h: Vertical displacement from the launch point (or ground, if launch from ground level).
- Conservation of mechanical energy: In the absence of non-conservative forces, total mechanical energy (kinetic + potential) remains constant.
Deriving the Height Using Kinematics
We model the motion with constant acceleration due to gravity g downward. Choose coordinates: launch from y = 0 with initial vertical velocity v sin θ and horizontal velocity v cos θ. At height h and speed 0.5v:
- Horizontal velocity remains v_x = v cos θ (no horizontal acceleration).
- Vertical velocity v_y at height h satisfies v_y^2 = (v sin θ)^2 − 2 g h.
- Speed condition: (v_x^2 + v_y^2) = (0.5 v)^2 = v^2 / 4.
Substitute v_x^2 = v^2 cos^2 θ and solve for v_y^2: v_y^2 = v^2 / 4 − v^2 cos^2 θ = v^2 (1/4 − cos^2 θ). Using the kinematic relation for v_y^2 and equating the two expressions for v_y^2 yields:
- (v sin θ)^2 − 2 g h = v^2 (1/4 − cos^2 θ).
Rearrange to isolate h:
- 2 g h = v^2 sin^2 θ − v^2 (1/4 − cos^2 θ) = v^2 (sin^2 θ + cos^2 θ − 1/4).
Since sin^2 θ + cos^2 θ = 1:
- 2 g h = v^2 (1 − 1/4) = (3/4) v^2.
Thus: h = (3 v^2) / (8 g). When expressed in terms of the launch angle, a more general form is h = (3 v^2 sin^2 θ) / (8 g), which reduces to the above for vertical launch (θ = 90°).
Step-by-Step Calculation Process
To solve for h in practice:
- Identify knowns: initial speed v, launch angle θ, and g ≈ 9.81 m/s².
- Compute vertical component: v_y0 = v sin θ.
- Use energy or kinematics to relate speed and height: v^2/4 = v_x^2 + v_y^2, with v_x = v cos θ.
- Solve for v_y^2 = v^2 (1/4 − cos^2 θ) and equate with v_y0^2 − 2 g h.
- Rearrange to h = (v^2 sin^2 θ − v^2 (1/4 − cos^2 θ)) / (2 g) = (3 v^2 sin^2 θ) / (8 g).
- Plug numbers to compute h. Ensure the resulting h is physically reachable (0 ≤ h ≤ max height).
Numerical Example
Assume v = 20 m/s and θ = 90° (straight up). Then:
- v^2 = 400 m²/s².
- h = (3 × 400) / (8 × 9.81) ≈ 1200 / 78.48 ≈ 15.3 m.
The maximum height for this throw is H = v^2 / (2 g) ≈ 20.4 m, so 0.5v occurs well before the peak, on the way up. For θ = 30° and the same v:
- sin 30° = 0.5, sin^2 30° = 0.25.
- h = (3 × 400 × 0.25) / (8 × 9.81) ≈ 3.8 m.
These examples illustrate how h scales with v^2 and sin^2 θ.
Physical Interpretation and Trajectory Context
Speed decreases as the projectile rises, reaches a minimum at the peak (if not vertical, the horizontal component remains), then increases on the way down. The condition speed = 0.5v can occur twice on an up-and-down trajectory if 0.5v is greater than the speed at the peak. For vertical launches, there is one height on the ascent and one on the descent where speed equals 0.5v, symmetric in speed but not necessarily in time. For shallower angles, the horizontal component dominates, so the speed may never drop to 0.5v if v cos θ > 0.5v (i.e., cos θ > 0.5, θ
Comparison of Equivalent Forms
| Quantity | Formula | Context |
|---|---|---|
| Height for speed = 0.5v (vertical launch) | h = 3 v^2 / (8 g) | Simplest case; directly proportional to v^2 and inversely proportional to g. |
| Height for speed = 0.5v (general angle θ) | h = (3 v^2 sin^2 θ) / (8 g) | Scales with the squared vertical component; zero for θ = 0°. |
| Maximum height | H = v^2 sin^2 θ / (2 g) | Useful for checking if 0.5v is reachable: h ≤ H. |
| Speed at any height h | v_h = sqrt(v^2 − 2 g h) |
Common Misconceptions and Pitfalls
- Ignoring horizontal velocity: Speed is not just vertical speed; always use v^2 = v_x^2 + v_y^2.
- Assuming one height for all angles: The height depends strongly on θ; for shallower angles, the required height can be much smaller or nonexistent.
- Forgetting domain limits: Confirm that the solved h lies between 0 and the maximum height; negative or excessive h indicate no physical solution.
- Using total energy incorrectly: In energy methods, remember that kinetic energy at height h is (1/2)m(0.5v)^2, and potential energy is mgh.
When No Real Solution Exists
At angles close to horizontal (θ → 0°), v cos θ ≈ v, so speed remains near v and never reaches 0.5v. Mathematically, requiring v^2 (1/4 − cos^2 θ) ≥ 0 for real v_y implies cos^2 θ ≤ 1/4, i.e., |cos θ| ≤ 0.5, which corresponds to θ ≥ 60° from the horizontal (or θ ≤ −60°). For |θ| 60°, there are two heights (ascending and descending) where speed equals 0.5v; for |θ| = 90°, there is one height at the single point where speed passes through 0.5v.
Units, Scaling, and Practical Tips
- Use consistent units: if v is in m/s and g in m/s², h comes out in meters.
- Height scales with the square of initial speed: doubling v quadruples h for a given θ.
- On other planets (different g), scale h inversely with g; for example, on the Moon (g ≈ 1.63 m/s²), heights for the same v and θ are about 6 times larger.
- When solving problems, first check angle feasibility (θ ≥ 60° for real solutions) before plugging into formulas.
Summary and Takeaways
The height at which a projectile’s speed equals half its initial speed is given by h = (3 v^2 sin^2 θ) / (8 g), derived from kinematics or conservation of energy. The solution exists only when the launch angle is steep enough (|θ| ≥ 60°) to allow the speed to decrease to 0.5v. The height is proportional to v^2 and sin^2 θ, and inversely proportional to g. Always verify that the computed height lies between 0 and the maximum height of the trajectory to ensure physical realism.