Find theta given range, height, and projectile velocity

[rileybrady]

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!

 

[anathema]

Hello, it seems like you are on the right track in terms of setting up the problem and using the equations correctly. However, one thing to note is that the vertical distance (d=30m) should be set equal to the initial vertical velocity (Voy) multiplied by the time (t) plus half of the acceleration due to gravity (1/2at^2). This is because the projectile's initial vertical velocity (Voy) is not zero, it is equal to 50*sinx.

So the correct equation for the vertical distance would be:

d=Voy*t+1/2at^2

Plugging in the values we have:

30=50*sinx*t+1/2(-9.8)(220/50cosx)^2

Now, since we already have t in terms of x from the horizontal equation, we can substitute that into the vertical equation and solve for x.

30=50sinx*(220/50cosx)+1/2(-9.8)(220/50cosx)^2

Solving for x, we get:

x=arcsin(30/((220/50)^2+9.8(220/50)^2))

x=arcsin(30/240.4)

x=arcsin(0.1246)

x=0.125 radians or 7.17 degrees.

So, the angle of launch for the projectile to just make it over the wall would be approximately 7.17 degrees. Hope this helps!