Polar vector to rectangular equation

[x14ace]

Homework Statement

The problem I'm stuck on is to come up with an equation which will model the path of the projectile in an xy-plane.
We are given:
r(t)={ [(v₀cos(θ))t], [h+(v₀sin(θ))t-(½)gt²] }
v₀=100 ft/s
θ=60°
h=4 ft
so basically,
r(t)={[100cos(60)t], [4+100sin(60)t-16t²]}

Homework Equations

I know that:
x=rcosθ
y=rsinθ
tanθ=y/x

The Attempt at a Solution

I tried to get the t out for both components.
x=100cos(60)t
x=50t
t=x/50

y=4+100sin(60)t-16t²
y=4+86.6t-16t²
I used the quadratic equation and ended up with
t(0)=(-86.6±88.1)/8 =-21.83, 0.1875
I thought that when I solved for the t's and set them equal to each other, I would get an equation in xy coordinates but I'm not sure where to go from here.

I also tried to use x=rcosθ
since x=50t,
50t=rcos(60)
r=100t
x²+y²=r²
so (50t)²+y²=(100t)²
y²ends up equalling 7500t²
and I'm back to being stuck.

 

[BvU]

Hi ace, welcome to PF !
Seems to me you mistake r(t) for a vector in polar coordinates $\vec r(t) =(r(t),\theta(t))$.
It is given to you already in cartesian (rectangular) coordinates: $\vec r(t) = (x(t), y(t))$

My guess is you now need to convert to y(x) to draw the trajectory (but I can't conclude that from the problem statement -- could also be the exercise wants $(r(t),\theta(t))$

[edit] when I read again, I think you are well underway writing $x(t) = v_0\cos\theta t \Rightarrow t = x/(v_0\cos\theta)$ and substitute that t in the expression for y.

 

[SteamKing]

Homework Statement

The problem I'm stuck on is to come up with an equation which will model the path of the projectile in an xy-plane.
We are given:
r(t)={ [(v₀cos(θ))t], [h+(v₀sin(θ))t-(½)gt²] }
v₀=100 ft/s
θ=60°
h=4 ft
so basically,
r(t)={[100cos(60)t], [4+100sin(60)t-16t²]}

Homework Equations

I know that:
x=rcosθ
y=rsinθ
tanθ=y/x

These equations in Section 2 are unnecessary.

The Attempt at a Solution

I tried to get the t out for both components.

Why? Don't you think the path of a projectile depends on the time variable t?

This is an example of what is known as a parametric equation, one where the values of x and y depend on, or are expressed, in terms of another variable called a parameter. In this case, the parameter is time, or t. Sometimes the parameter can be eliminated, but this is not always necessary nor desirable.

x=100cos(60)t
x=50t
t=x/50

y=4+100sin(60)t-16t²
y=4+86.6t-16t²
I used the quadratic equation and ended up with
t(0)=(-86.6±88.1)/8 =-21.83, 0.1875
I thought that when I solved for the t's and set them equal to each other, I would get an equation in xy coordinates but I'm not sure where to go from here.

I also tried to use x=rcosθ
since x=50t,
50t=rcos(60)
r=100t
x²+y²=r²
so (50t)²+y²=(100t)²
y²ends up equalling 7500t²
and I'm back to being stuck.

You had your answer back at the beginning:

r(t)={[100cos(60)t], [4+100sin(60)t-16t²]}

All you needed to do was evaluate the terms of r(t) for which you were given values (like h and θ) and express r(t) in terms of t.

x(t) = 50t
y(t) = 4 + 86.6t-16t²

If you want to plot r(t), all you need to do is substitute different values of the parameter t and you can determine x(t) and y(t) from those.