Projectile Motion
Split the launch velocity into horizontal and vertical components, then run constant velocity across and constant acceleration downwards. For a same-height flight with gravity alone and a stated g, range, greatest height and flight time follow.
What a learner can do afterwards
- Split a launch speed and angle into horizontal and vertical components
- Find the time of flight, range, maximum height, and speed from the vertical and horizontal motions
- Say what the model assumes by leaving air resistance out
1 · Read
A projectile in flight feels gravity alone. Air resistance is ignored, so downward acceleration is constant. Treat g as a positive magnitude in metres per second squared, with ay = -g. Take right and up as positive, and measure theta above the positive horizontal axis. A launch speed u splits into ux = u cos(theta) across and uy = u sin(theta) up. The launch position is (x0, y0), and y = 0 is the level landing ground in the examples.
The horizontal life is constant velocity: x = x0 + ux t. The vertical life is constant acceleration: y = y0 + uy t minus g t squared over 2, and vy = uy minus g t. The two equations share the elapsed time t and the launch parameters. Horizontal velocity stays at its initial value, which may be zero. At a reached apex vy is zero, while the downward acceleration g stays nonzero.
A ball launches from ground level at 25 m/s with cos(theta) = 0.8 and sin(theta) = 0.6. The components are ux = 20 m/s and uy = 15 m/s. With g = 10, the flight time back to launch height is 2uy/g = 30/10 = 3 s. The range is ux times t = 20 times 3 = 60 m. The maximum rise above launch is uy squared over 2g = 225/20 = 11.25 m.
The launch splits into 20 m/s across and 15 m/s up.
The formula 2uy/g gives the flight time only when the launch is upward and the landing returns to the launch height. A different landing height needs the vertical position equation solved as a quadratic. The maximum rise uy squared over 2g is measured above the launch level and needs an upward launch with the apex reached before landing. Add the initial height y0 for the elevation above ground.
A ball launches from a 15 m platform at 10 m/s vertically upward. Take g = 10. The vertical position is y = 15 + 10t minus 5t squared. Setting y = 0 gives 5t squared minus 10t minus 15 = 0, which is t squared minus 2t minus 3 = 0, factoring as (t minus 3)(t plus 1) = 0. The future positive root is t = 3 s.
Speed at any instant combines the components: speed = the square root of vx squared plus vy squared. Horizontal velocity stays at its initial value throughout, while vy = uy minus g t. At a reached apex vy is zero, so the speed there equals the magnitude of ux. A vertical launch has ux = 0, so its apex speed is zero.
Split the launch into ux and uy, run x and y with the same elapsed time, then read range, flight time and height from the vertical life.
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