# 2d motion range , find angle (Really simple, not sure what I'm missing)

1. Sep 28, 2014

### Chas3down

1. The problem statement, all variables and given/known data
Firing something at 40m/s at angle X, it lands 65meters away, give 2 angles it could be.. (Assume g=10m/s^2)(Fired at y=0)(No air resistance)

2. Relevant equations

D = v^2/g sin(2x)

3. The attempt at a solution

d= v^2/g sin(2x)

65 = 40^2/10 sin(2x)

650/1600 = sin(2x)

13/32 = sin(2x)

arcsin(13/32)/2 = x
23.97/2 = x

x = 11.985
and
x = 90-11.985

But it's wrong?

Heres another form it could be in, but don't know where to go from here..
13/32 = 2sinxcosx
13/64 = sinxcosx

2. Sep 28, 2014

### Orodruin

Staff Emeritus

3. Sep 28, 2014

### Chas3down

There isn't one, it just says if it is correct or not (anwser must be exact)

With further research, it seems one angle is
"arcsin(13/32)/2" , which is what I had but in decimal form, however can't find the second angle.. , i tried 90 - arcsin(13/32)/2, 180 - arcsin(13/32)/2, 360 - arcsin(13/32)/2, - arcsin(13/32)/2, and arcsin(-13/32)/2, and arccos(13/32)/2

"90 - arcsin(13/32)/2" and "arccos(13/32)/2" work, but I am just not sure what form they want it in, probably a manipulation of 13/32 so the form would be arcsin(??)/2 for second angle

Last edited: Sep 28, 2014
4. Sep 28, 2014

### Orodruin

Staff Emeritus
This question seemed to indicate that you had compared it with a solutions manual or equivalent which had stated something else.

You can argue for this either completely through mathematics - or by physical reasoning: If the angle would be 180 - 12 degrees, would the projectile land in front of the firing point?

The mathematical approach is to note that sin(2x) = a implies that 2x = asin(a) or 2x = 180o - asin(a).