Question

Im stuck on a physics/calculus problem... if someone can help me that would be greatly appreciated :)

An arrow is shot into the air and moves along the parabolic path y = x(50-x). The horizontal component of velocity is always 30 ft/s. What is the vertical component of velocity when (i) x = 10 and (ii) x = 40?

Answer 1

I think the answer is 400, but im not sure...

Answer 2

well, I haven´t done the calculations, but the problem is quite simple.
As you can see from the equation, the range of the arrow is 50ft, and the arrow reaches the maximum point of the parabolic path at the medium point of its range, so the time it takes to go from x=0 to x=25 is: \[t_{mid}=\frac{25ft}{30ft/s}\]
now, at the maximum point, the vertical component of the velocity will be null, so if you multiply the time calculated by the gravity, you have que initial vertical component of the velocity:
\[V^{0}_{vertical} = g \times t_{mid}\]
and now, you can calculate the time that it takes to move the arrow from x=0 to x=10 and the vertical component of velocity at this time will be given by:
\[V_{vertical}=V^{0}_{vertical} - g\times t_{x=10} \]
and, by simetry, the component at x=40 will have the same magnitude, but the oposite direction

Answer 3

Should be 120.27 ft/s, at x=10ft, y=400ft, also your horizontal velocity is 4.01ft/s and your vertical velocity at t=0 is 200.47ft/s


Check this out
http://www.wolframalpha.com/input/?i=projectile+motion&a=*FS-_**Projectile.t-.*Projectile.v-.*Projectile.alpha--&f2=200.510+ft%2Fs&f=Projectile.v_200.510+ft%2Fs&f3=88.85&x=10&y=8&f=Projectile.alpha_88.85&a=*FVarOpt.1-_**-.***Projectile.m---.*--

and your original question
http://www.wolframalpha.com/input/?i=y%3Dx%2850-x%29

Answer 4

Thanks!