Projectile motion

Matematisk analys av kaströrelse

Equations of motion: no air resistance
We first consider the situation of a projectile launched from a tower of height h onto some impact function , ignoring the e↵ect of air resistance. In order to solve for ✓m, we need to find equations for motion in the x- and y-directions. We define ✓ to be the angle above the horizontal at which the projectile is launched. The projectile is launched with an initial velocity v, which has magnitude v, and when broken up into x- and y-components, gives us the initial conditions
x(0) = 0; y(0) = h; x0(0) = vcos✓; y0(0) = vsin✓.
Without air resistance, acceleration in the x-direction is zero, while in the y-direction it is solely due to gravity, where g = 9.8 m/s2. Thus we can solve the second-order di↵erential equations to find our two motion equations. In each step, we integrate both sides of the
5
Vertical distance
equation with respect to t and apply our initial conditions. In the x-direction, we have
x00(t) = 0; x0(t) =vcos✓;
x(t) =vtcos✓. (1) The motion in the y-direction is described by
y00(t) = g;
y0(t) =gt+vsin✓;
y(t) =1gt2+vtsin✓+h. (2) 2
We now have a set of parametric equations for the motion of the projectile as a function of t, but to maximize the projectile’s horizontal distance, we want to find a path function, p, that defines the projectile’s height as a function of horizontal distance, x. Solving for t in (1) and substituting into (2) yields
and therefore
t=x, vcos✓
p(x)=h+vsin✓⇣ x ⌘1g⇣ x ⌘2 vcos✓ 2 vcos✓
gx2 2
= h + x tan ✓ 2v2 sec ✓. (3)
We now have one equation that describes the motion of the projectile, which is useful in finding the launch angle that maximizes x.

Om mattelararen

Licentiate of Philosophy in atomic Physics Master of Science in Physics
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