--- title: "Projectiles Launched at an Angle" description: "With constant downward gravity and no air resistance, an angled launch gives horizontal and vertical starting velocities. Use vertical motion for flight time and greatest rise, then horizontal motion " canonical: https://lightmysky.com/learn/science/projectiles-launched-at-an-angle-mt_MMuacyCqb8 source: https://lightmysky.com/learn/science/projectiles-launched-at-an-angle-mt_MMuacyCqb8.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Projectiles Launched at an Angle With constant downward gravity and no air resistance, an angled launch gives horizontal and vertical starting velocities. Use vertical motion for flight time and greatest rise, then horizontal motion for range. Equal climb and fall times apply when landing is at launch height. Subject: Science · Area: Forces & Motion · Ages 16 to 18 Page: https://lightmysky.com/learn/science/projectiles-launched-at-an-angle-mt_MMuacyCqb8 ## Ready when they can - Resolves the launch velocity, then gets the time of flight from the vertical motion alone - Finds the range from the horizontal component and that time, without quoting a range formula - Explains why, at a fixed launch speed with constant downward gravity and no air resistance, 45 degrees gives the greatest range for equal-height landing, but a lower angle is best for landing on horizontal ground below a cliff ## Lesson: Launching at an angle You have used separate horizontal and vertical motion. Assume only constant downward gravity acts, with no air resistance. Let u be launch speed and theta the angle above horizontal. The starting components are u_x = u cos(theta) horizontally and u_y = u sin(theta) upward. For 30 degrees, sin(theta) = 0.5. Horizontal acceleration is zero, so u_x stays constant. With upward positive, vertical acceleration is -g. *(drawing: Resolve a 16 m/s launch at 30 degrees above horizontal. The cells show the calculation, not the flight path.)* First find u_y = u sin(theta), with theta measured above horizontal. For 30 degrees, sin(theta) = 0.5. At the top, vertical velocity is zero: 0 = u_y - gt, so climb time is u_y/g. Under constant downward gravity without air resistance, the fall back to launch height takes as long as the climb. Add climb and fall for total time T. Horizontal velocity stays u_x, so range R = u_x T. **Example.** For a 16 m/s launch at 30 degrees, u_y = 8 m/s and u_x is about 13.9 m/s. Ignore air resistance and take g = 10 m/s^2. Returning to launch height takes 2 times 8/10 = 1.6 s, giving range about 13.9 times 1.6 = 22.2 m. To find rise H above launch, set vertical velocity to zero in v_y squared = u_y squared - 2gH. Then H = u_y squared/(2g) = 8 squared/20 = 3.2 m. *(drawing: This line shows elapsed time. Equal launch and landing heights give equal climb and fall times.)* For fixed launch speed u, constant downward gravity, no air resistance and landing at launch height, flight time is 2u_y/g and range is 2u_x u_y/g. Since u_x squared + u_y squared = u squared and (u_x - u_y) squared is at least zero, 2u_x u_y cannot exceed u squared. Equality holds when u_x = u_y. The components are equal at 45 degrees above horizontal, so 45 degrees gives the greatest range. At fixed speed, with constant downward gravity and no air resistance, 45 degrees gives greatest range when landing at launch height. From a cliff top onto horizontal ground below, the best angle is below 45 degrees. The extra drop adds flight time, shifting the best balance toward horizontal velocity. Fall time exceeds climb time. Measure y from launch, upward positive. Use negative y for lower ground in y = u_y t - gt squared/2 to find time. **Recap.** Split the launch, take time from the vertical, and multiply sideways speed by that time. *Lesson, questions and flashcards written for this dataset.* ## Practice 8 questions on this page, each with its working shown. ## Needs first - [Projectile Motion as Two Independent Motions](https://lightmysky.com/learn/science/projectile-motion-as-two-independent-motions-mt_OBgL97RFbb) ## Optional connections - [Exact Values of Sine, Cosine and Tangent](https://lightmysky.com/learn/mathematics/exact-values-of-sine-cosine-and-tangent-mt_SNJEbXYOrR)