Why is the projectile motion of an object, in terms of vectors, r(t) = (u0cos(α))ti + [(u0sin(α)t-1/2gt^2)]j ?

Question

This is what my book says:

F=ma = -mgj where F and a are vectors and j is the vector component.
g= |a|, which is about 9.8m/s^2. Thus a=-gj
since v'(t) = a, we have v(t) = -gtj+C (t is the time variable and C is the position vector)
Where C=v(0)=v0 (v0, here, is a vector).
Therefore r'(t) = v(t) = -gtj + v0.
Integrating again, we obtain r(t) = -1/2gt^2j + tv0
If we write |v0| = u0 (the initial speed of the projectile, then
v0=u0cos(α)i+u0sin(α)j
and equation 3 becomes
r(t) = (u0cos(α))ti + [(u0sin(α)t-1/2gt^2)]j.

I can't see where they derived this from. Most of the book skips a lot of steps you you kind of just have to go over it over and over until you see why. I just can't see what I'm overlooking/not getting. Thanks in advance.

Oh, and just to clarify, α is the angle at which a particle is being thrown.

Pinkgreen · Accepted Answer

My opinion:
Let r be the position vector of the projectile at time t. Then
r"=-gj =>
r'=-gtj+r'(0), where r'(0) is the initial velocity=u(0)cosAi+
u(0)sinAj, where u(0)=the initial speed; A=the angle of
projection at t=0=>
r'=u(0)cosAi+[u(0)sinA-gt]j=>
r=u(0)cosAti+[u(0)sinAt-g(t^2)/2]j+r(0), 
r(0)=the initial position vector at t=0.

Elizabeth M · Answer

The C of line 3 is the initial velocity not a position vector as seen in line 4.
The equation for r(t) should have a constant which is zero if the projectile starts from
O .
 u0cos(Î±) and u0sin(Î±) are the initial horizontal and vertical components of V0
giving v0=u0cos(Î±)i+u0sin(Î±)j
Now sub this into r(t) = -1/2gt^2j + tv0 to give
r(t)=-1/2gt^2j + t[u0cos(Î±)i+u0sin(Î±)j]
    = -1/2gt^2j+u0cos(Î±)ti+u0sin(Î±)tj
   =u0cos(Î±)ti+(u0sin(Î±)t -1/2gt^2)j as given

Trending News

Why is the projectile motion of an object, in terms of vectors, r(t) = (u0cos(α))ti + [(u0sin(α)t-1/2gt^2)]j ?

2 Answers