Find theta given range, height, and projectile velocity

  • Thread starter rileybrady
  • Start date
  • #1

Homework Statement


If a trajectory is shot at angle theta at a fixed projectile velocity of 50 m/s, what angle(s) could it be shot so that it just makes it over a wall 30 meters high, 220 meters down the path.


Homework Equations



d=1/2at^2+vt
sinx*v=vertical component
cosx*v=horizontal component


The Attempt at a Solution



Time connects the horizontal and vertical distances, so my approach is to solve for time and then substitute.

On the vertical end, we can say:

Vo=50*sinx
a=-9.8
d=30m

Horizontal:

Vo=50*cosx
Vf=50*cosx
a=0
d=220

I solve for time on the horizontal time in terms of x.

d=vt
220=50cosx*t
t=220/50cosx


I can now use that t and plug it into the vertical t, ensuring that x is the only variable and solve for x.


d=1/2at^2+Vot
30=1/2(-9.8)(220/50cosx)^2+50sinx(220/50cosx)


After setting that up, I am stuck. I have tried it multiple times, combining them and using LCD, etc. I end up with a cosx and a sinx and can't use trig identities or anything to isolate x. I can't find any way to solve for the angle of launch



Thanks for the help!!
 

Answers and Replies

Related Threads on Find theta given range, height, and projectile velocity

Replies
1
Views
6K
Replies
8
Views
6K
Replies
2
Views
3K
Replies
4
Views
854
Replies
9
Views
14K
  • Last Post
Replies
3
Views
899
Replies
4
Views
26K
  • Last Post
Replies
5
Views
41K
  • Last Post
Replies
2
Views
1K
Top