Concept 09 / 10Kinematics
Projectile Motion: Trajectory, Time of Flight, Range and Maximum Height
The independence of horizontal and vertical motion, full derivation of the parabolic trajectory with time of flight, range and maximum height, complementary angles, and radius of curvature of the path.
A projectile is a body given an initial velocity and then left to gravity alone. The single idea that unlocks everything is the independence of perpendicular components: because has no horizontal component, the horizontal motion is uniform and the vertical motion is free fall, and neither influences the other. They share only one thing, the clock. So for a launch speed at angle above the horizontal, is constant forever while obeys .
Writing the two motions out: and . Eliminating gives the trajectory , a downward parabola. Because it is quadratic in with no cubic or higher terms, any trajectory equation of the form (with ) is a projectile, and you can read off and to recover the launch conditions.
The standard ground-to-ground results follow by setting . Time of flight: , so besides we get . Maximum height occurs when at , giving . Horizontal range . Since , the range is maximum at with ; and because , the complementary angles and give the same range (with different heights and flight times). A neat relation linking them: , so , and the two complementary flight times satisfy .
Two facts about the velocity deserve emphasis. At the highest point the velocity is purely horizontal with magnitude , which is the minimum speed of the whole flight; it is never zero unless the launch is vertical. And the acceleration is downward at every instant, including at the top; the top is where the velocity is momentarily horizontal, not where the body is momentarily free of gravity. A related Advanced favourite is the radius of curvature, obtained from where is the component of perpendicular to the velocity: at the top , and at launch .
For horizontal projection from a height with speed , the vertical motion starts from rest, so gives , independent of , and the horizontal range is . This independence is the content of the classic demonstration in which a ball rolled off a table and a ball dropped from the table edge strike the floor at the same instant.
Gotchas to hold onto. The formulae , , above assume launch and landing at the same level; for a projectile from a cliff you must solve the quadratic in afresh. For a projectile launched from a moving platform, add the platform velocity vectorially to get in the ground frame. And 'maximum range' at holds only for launch and landing at the same height; from a height the optimal angle is less than .
Key Takeaways:
- Components: , , constant, .
- Trajectory: .
- Same-level results: , , , at .
- ; complementary angles give equal ranges; .
- Horizontal projection from height : (independent of ), .
- Radius of curvature: where is the angle of the velocity with the horizontal; at the top .
- Pitfall: using when the landing level differs from the launch level.
- Pitfall: thinking the speed or the acceleration is zero at the top.
Brain-Teaser 1: A projectile's trajectory is with in metres and m/s. Find the launch angle, launch speed, range, maximum height and time of flight.
Solution: Compare with . Matching the linear term, , so . Matching the quadratic term, , so with we get , giving and m/s. Range: set , so , giving m, which matches m. Maximum height is at m, where m, matching m. Time of flight s. Every quantity is extractable from the trajectory alone, with no extra data.
Brain-Teaser 2: A hunter aims a rifle directly at a monkey hanging from a branch. At the exact instant the rifle is fired, the monkey lets go and falls. Ignoring air resistance, does the bullet hit the monkey, and does the answer depend on the muzzle speed?
Solution: Work in the monkey's frame. Both bullet and monkey have acceleration downward, so and the bullet moves relative to the monkey in a straight line with its initial relative velocity, which points exactly along the original line of sight. Therefore the bullet always heads straight for the monkey and hits, whatever the muzzle speed. Equivalently, in the ground frame, both drop by the same from where they would have been in the absence of gravity, and the aim was correct in that gravity-free world. The muzzle speed does matter for one thing: it determines how far the bullet travels before it hits, so a slow enough bullet will reach the monkey's horizontal position only after the monkey has hit the ground, in which case the hit fails for geometric reasons, not kinematic ones. The classic wrong answer, 'the hunter should aim above the monkey to compensate for the drop', is exactly wrong when the target also falls.
Brain-Teaser 3: A ball is projected with m/s at to the horizontal, m/s. Compare the radius of curvature of its path at the launch point and at the highest point, and explain why they differ so much.
Solution: The radius of curvature satisfies , where is the component of perpendicular to . At the top, is horizontal with m/s, and the whole of is perpendicular to it, so m. At launch, m/s and the velocity makes with the horizontal, so the perpendicular component of gravity is m/s, giving m. The path is eight times straighter at launch. The reason is twofold: the speed is larger at launch (and grows as ), and only part of gravity acts to bend the path there, the rest merely slowing the body down. At the top nothing is being slowed, so all of gravity goes into turning, and the path is at its most sharply curved.
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