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Projectile motion

A projectile runs two problems at once: constant velocity across, constant acceleration down, and the two never interfere. Time is the only thing they share, and it is the bridge every projectile calculation crosses.

Year 12AQA 3.4.1.4

Builds on Motion graphs and the SUVAT equations.

IN THIS TOPIC

  • Explain the independence of horizontal and vertical motion in a uniform gravitational field.
  • Solve projectile problems by treating the two directions separately, linked only by time.
  • Describe qualitatively how air resistance changes a projectile's trajectory.

WHAT YOU PROBABLY THINK

Something must keep pushing a projectile forwards.

Two motions, one clock

Once a projectile leaves your hand, ignoring air resistance, exactly one force acts on it: its weight, straight down. Nothing pushes it forward, and nothing needs to; motion continues without a force. The consequence is the whole topic: the horizontal and vertical motions are independent.

Dropped and launched at the same instant: identical heights at every momentdroppedlaunched sideways
FIG. 1One ball dropped, one launched sideways at the same instant. At every moment they are at exactly the same height: the sideways motion changes the vertical story not at all.

Drop one ball and launch another horizontally at the same instant, and they hit the ground together. The launched ball's sideways speed does nothing to its fall. Horizontally it moves at constant velocity; vertically it accelerates at g like anything else in free fall.

Solving the two problems

Every projectile question is the same procedure. Resolve the initial velocity into components. Treat the vertical motion as constant acceleration, a = g downward, using the SUVAT equations. Treat the horizontal motion as constant speed, distance = speed × time. The single shared quantity is time, so almost every problem finds t from one direction and spends it in the other.

Velocity components along a projectile's path: horizontal constant, vertical growinghorizontal component: unchangedvertical component: grows
FIG. 2Velocity components along the flight. The horizontal component never changes; the vertical one shrinks to zero at the top and grows again coming down.

The velocity picture follows: the horizontal component is identical at every point of the flight, while the vertical component passes through zero at the apex. At the top of the arc the projectile is still moving, horizontally, which is why “velocity at the highest point” has a non-zero answer.

What air resistance does

AQA asks for the qualitative story. Air resistance opposes motion and grows with speed, so it steals horizontal speed throughout the flight and fights the vertical motion both ways, up and down.

Air resistance makes the real trajectory lower, shorter and steeper on the way downno air resistancewith air resistance
FIG. 3The same launch with and without air resistance. The real path peaks lower, lands shorter and comes down more steeply than it went up.

The trajectory that results is lower and shorter than the ideal parabola, and it loses its symmetry: the descent is steeper than the ascent, because by then the horizontal speed has been eaten away. The same reasoning caps a vehicle's top speed: as speed rises, resistive forces rise to meet the driving force, and the maximum speed is where they balance.

THE EXAM BIT

  • Start by resolving the launch velocity and writing two columns, horizontal and vertical. Mixing components in one equation is the topic's defining error.
  • The horizontal direction has no acceleration: never put g in it, and never use SUVAT there; distance is simply speed times time.
  • At the apex the vertical velocity is zero but the speed is not: the horizontal component survives. Both facts get asked.
  • Time links the two directions. Found from the vertical drop, it plugs straight into the horizontal range, and vice versa.
  • For air resistance, three phrases score: lower maximum height, shorter range, descent steeper than ascent.

CHECK YOURSELF

A ball rolls off a table 1.25 m high at 4.0 m s−1. How long is it in the air, and how far from the table does it land? Take g = 9.8 m s−2.

Show a hint

The fall time comes from the vertical motion alone; the sideways speed is irrelevant to it.

Show the answer

Vertical: s = 12gt2 with s = 1.25 m gives t2 = 2 × 1.25 / 9.8 = 0.255, so t = 0.50 s.

Horizontal: constant 4.0 m s−1 for 0.50 s gives a range of 4.0 × 0.50 = 2.0 m.

Notice the order: the vertical problem produced the time, and the horizontal problem spent it.

Across: constant velocity.

Down: constant acceleration. Time joins them.

No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.