Question

Show that the trajectory of a particle that is launched at an angle is parabolic?

Answer 1

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

`{: (v=u+at, " where ", s="displacement "(m)), (s=ut+1/2at^2, , u="initial speed "(ms^-1)), (s=1/2(u+v)t, , v="final speed "(ms^-1)), (v^2=u^2+2as, , a="acceleration "(ms^-2)), (s=vt-1/2at^2, , t="time "(s)) :}`

Consider a projectile launched at an angle `theta` with initial speed `u`. We ignore air resistance, and model the projectile as a particle with all mass concentrated at a point.

Horizontal Motion

The projectile will move under constant speed (NB we can still use "suvat" equation with a=0).

The projectile will travel a distance `x` in time `t`, we must resolve the initial speed in the horizontal direction

`{ (s=,x,m),(u=,u cos theta,ms^-1),(v=,"Not Required",ms^-1),(a=,0,ms^-2),(t=,t,s) :}`

So applying `s=ut+1/2at^2` we get

`x = u \ costheta \ t`

Vertical Motion

The projectile travels under constant acceleration due to gravity. Considering upwards as positive, and again resolving the initial speed (vertically this time):

`{ (s=,y,m),(u=,u sin theta,ms^-1),(v=,"Not Required",ms^-1),(a=,-g,ms^-2),(t=,t,s) :}`

Applying `s=ut+1/2at^2` we have:

`y = u \ sin theta \ t+1/2(-g)t^2`
`\ \ = u \ sin theta \ t - 1/2 g t^2`

Substituting for `t` from the first equation into the second we get:

`y = u \ sin theta \ (x/(u cos theta)) - 1/2 g (x/(u cos theta))^2`
`\ \ = (tan theta) x - g/(2u^2 cos^2 theta) \ x^2`
`\ \ = a x - b x^2 \ \`, say, where `a,b` are constants.

Which is the trajectory of a parabola.