Need a quick physics refresher...

[b0mbrman]

It's been many years since I had to do a problem like this...how close am I to being correct?

Let's say I'm launching a projectile with initial velocity V, initial height of zero, at an angle of A° (Where 0<A<90) and I want to know how far the cat will travel.

First step is to break V into two parts--horizontal (Vx) and vertical (Vy)

Vx is constant and is cos(A) * V
Vy starts at sin (A) * V and is constantly decreasing at a rate of -9.8 (m/s)/s

Uh...so yeah, what's the answer here for the distance in terms of V and A?

 

[BornStar]

Um, well, this is hardly simplified, but I think its right. x = V cos(A) * ((V sin(A)) / 9.8) + (4.9 * V^2 * sin^2 (A)) / 9.8^2

x is distance in meters, A is the angle and V is the velocity in m/s

Please, anybody feel free to correct me if there are any mistakes.

 

[dullard]

<< how far the cat will travel >>

:| That is quite a terrible thing to say. Lets all see how far b)mbrman's computers will travel.

You have the basics down alright for a rough estimate for a launch from sea level on Earth. However that cat might be affected by wind speed and air resistance. You don't have those in your model, so you won't be as accurate as you could be.

In your model the change in velocity only depends on gravity:
dv/dt = -g

Integrate once: v = v(0) - gt
where v(0) = sin(A)*V

The change in height is equal to the velocity:
dy/dt = v = v(0) - gt

Integrate once: y = y(0) + v(0)t - gt^2/2

At what time will the cat land (denote this time as T)?
0 = y(0) + v(0)*T - gT^2/2

Solve the quadratic equation:
T = v(0)/g + [V(0)^2/g^2 +2y(0)/g]^.5

Suppose you launched your cat from ground height - this will simplify since y(0)=0:
T = 2v(0)/g

Where is the cat's horizontal position?
dx/dt = Vx = cos(A)V

Integrate once: x = x(0) + cos(A)Vt

The cat will land at t=T so:
x= x(0) + 2cos(A)v(0)V/g = x(0) +2cos(A)sin(A)V^2/g

The distance traveled is x - x(0) = 2cos(A)sin(A)V^2/g

The math may look terrible since this is done in text. Now let me check for typo's.