This is repetition from the previous page
Projectiles
x = u cosø; y = ut sinø – ½ gt².
Substituting to eliminate t gives the ‘big’ equation,
y = x tanø – g x² (1 + tan²ø) / 2u². This is quadratic in both tanø and in x. There are some fairly hard questions that require you to be very clear which quadratic you should be working on, such as those for maximum range under constraints.
Range, from the firing point back to the same vertical level, is R = u².sin2ø /g and the maximum height is given by H = u.sin²ø /2g. You need the big equation to solve cases where the impact is not at the same height as the firing point. There are some ‘nice’ equations formed by expressing R and H together, often found in extension papers.
Problems are made complicated by having two projectiles in flight, or by firing up a slope; both of these cases are solved by setting up simultaneous equations to reflect the motion.
Whenever you are having difficulty, separate the horizontal and vertical components and take care to look at where things are at a point in time.
For further maths, air resistance is added. The manageable version has this as F = mkv and the more difficult one mkv2. The handling of the differential equations is as at the top of this page, and good practice. Modelling something like a tennis serve is valid for rubbish player with only the gravity: for class players, only air resistance is sufficient. A good investigation to do.
1. A small projectile is required to land on the floor of a disant open box, length a and height b. Write a condition for the minimum angle of projection.
Might this be also written as a tanø = Max Range ?
Far too many projectiles examples basically demonstrate armed conflict for this politically correct world we live in. Here are a few such:
2. A battle tank is firing up a long slope at 15º. Muzzle velocity is around 300m/s. Ignoring air resistance and ignoring the height of the gun, what is the maximum expected range upslope and the angle to the slope 𝞪 required to produce this?
show that sin (2†-𝞪)-sin𝞪 = hard to type
2. Rmax = u²/g (1-sin 15)/cos² 15
case for projection upslope requires proof of trig identity to solve. And typing it is hampered by not only the difficulty of index, but also the greek symbols, Dammit.
