Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — time of flight, range and greatest height
A ball is projected at 30 m s−1 at 40° to the horizontal from level ground. Find the time of flight, the range and the greatest height. Take g=9.81.
- 1ux=30cos40°=22.98 and uy=30sin40°=19.28 m s−1Split the velocity first, always.
- 2Vertically, y=0 on landing: t(uy−21gt)=0t=0 is the launch.
- 3T=g2uy=9.812(19.28)=3.93 s
- 4R=uxT=22.98(3.93)=90.3 mHorizontal velocity is constant, so distance is just speed × time.
- 5H=2guy2=19.62371.7=18.95 mReached at t=1.97 s, half the flight time.
Example 2 — clearing an obstacle using the path equation
For the same projection, find the height of the ball when it is 20 m horizontally from the launch point.
- 1No time is mentioned, so use the trajectory equation.That is the signal to eliminate t.
- 2y=xtanα−2u2cos2αgx2
- 3=20tan40°−2(900)cos240°9.81(400)
- 4=16.78−1056.33924=16.78−3.72
- 5y=13.1 mBelow the maximum of 18.95 m ✓ — it has not yet reached the apex at x=45.2 m.
Example 3 — projection from a height
A ball is thrown horizontally at 15 m s−1 from a cliff 20 m high. Find the time to land, the horizontal distance and the landing speed.
- 1Horizontal projection means uy=0.Vertically it is a free drop.
- 2Taking downwards positive: 20=21(9.81)t2Do NOT use T=2usinα/g — the ground is not level with the launch.
- 3t2=4.077, so t=2.02 s
- 4x=15(2.02)=30.3 mHorizontal speed never changes.
- 5vy=gt=9.81(2.02)=19.81 m s−1
- 6Speed =152+19.812=24.8 m s−1, at 52.9° below the horizontalRecombine the components at the end.