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.
What a learner can do afterwards
- 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
1 · Read
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.
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.
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.
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.
Split the launch, take time from the vertical, and multiply sideways speed by that time.
Lesson, questions and flashcards written for this dataset.
Three readings and six flashcards written for this dataset. Linked video: Parametric equations for projectile motion, Khan Academy.
More to read
Choosing the starting components
Choose horizontal and upward directions before calculating a landing point. Measure the launch angle from horizontal. A speed of 14 m/s at 30 degrees gives an upward component of 14 times 0.5 = 7 m/s. The horizontal component is 14 cos(30 degrees), about 12.1 m/s.
Use each component in its own direction. Putting 14 m/s into both directions would describe a different launch.
Assume gravity is constant and downward, and ignore air resistance. There is then no horizontal acceleration. The horizontal component remains about 12.1 m/s while the upward velocity decreases. At the peak, vertical velocity is zero; horizontal motion continues.
Both calculations use the same elapsed time. Find when the ball reaches its landing height using vertical motion. Then multiply that time by the horizontal component to find range.
Finding time, range and rise
A ball starts with components of 8 m/s horizontally and 15 m/s upward. It returns to its launch height. Ignore air resistance and use constant downward g = 10 m/s^2.
The upward component reaches zero after 15/10 = 1.5 s. The fall back to launch height takes another 1.5 s. Total flight time is 3 s. Horizontal velocity remains 8 m/s, so range is 8 times 3 = 24 m.
For the rise H above launch, use the vertical equation at the peak: 0 = 15 squared - 2 times 10 times H. This gives H = 225/20 = 11.25 m. It measures rise above launch, not necessarily height above the ground.
When comparing launch angles, hold speed fixed. For equal-height landing in this model, range depends on the product of the initial components. That product is greatest when the components are equal, at 45 degrees above horizontal.
Allowing for a lower landing point
A ball starts 8 m above horizontal ground, with components of 9 m/s horizontally and 6 m/s upward. Ignore air resistance and use constant downward g = 10 m/s^2.
Take upward as positive and measure vertical displacement y from launch. The ground is at y = -8 m. The vertical equation becomes -8 = 6t - 5t squared. Rearranging and factoring gives (5t + 4)(t - 2) = 0. The solutions are -0.8 s and 2 s. Only 2 s is after launch, so the flight lasts 2 s. Range is 9 times 2 = 18 m.
The climb takes 6/10 = 0.6 s. Doubling it would give 1.2 s, when the ball has only returned to launch height. It still has 8 m to fall.
For fixed launch speed and cliff height in this ideal model, the greatest range onto the lower horizontal ground occurs below 45 degrees. The extra drop shifts the best balance toward horizontal velocity.
Explore more
2 · Watch
Take it off screen
Where it sits
Learn first
Optional connections
This opens up
Nothing builds on it yet.
Where this leads
Jobs that lean on this skill. Follow one to see everything it is built on.
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.