## 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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