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## 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!!