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

[Chas3down]

Homework Statement

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)

Homework Equations

D = v^2/g sin(2x)
[/B]

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

 

[Orodruin]

Your approach looks reasonable. What is the given answer?

 

[Chas3down]

Your approach looks reasonable. What is the given answer?

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

 

[Orodruin]

But it's wrong?

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

But I can't seem to get the second angle, i tried 180 - that and 90 - that

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 = 180^o - asin(a).

 

[HalfLostAndSearching]

I would first like to clarify the question and make sure all the necessary information is provided. It is unclear what the 2D motion range is referring to and what the goal of finding the angle is. Is the angle being asked for the initial launch angle or the angle at which the object lands? It is also unclear what the given 2 angles are supposed to represent. Are they the possible angles at which the object could be launched to land at a distance of 65 meters? Additionally, more information is needed about the object being fired, such as its mass and shape, in order to accurately calculate its trajectory.

Assuming that the goal is to find the initial launch angle at which the object will land at a distance of 65 meters, and that the two angles given are the possible launch angles, I would approach the problem by using the given equation d = v^2/g sin(2x) to solve for the two possible angles. However, I would first convert the given velocity of 40 m/s into its components in the x and y direction, as the object will follow a parabolic path in both directions.

Using the given values, the equation becomes:

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

Simplifying, we get:

13/8 = cos(x)^2 sin(2x)

Using the trigonometric identity sin(2x) = 2sin(x)cos(x), the equation can be rewritten as:

13/8 = cos(x)^2 (2sin(x)cos(x))

Dividing both sides by cos(x)^2, we get:

13/8cos(x)^2 = 2sin(x)cos(x)

Simplifying further, we get:

13/16 = sin(x)

Taking the inverse sine of both sides, we get:

arcsin(13/16) = x

x = 55.89° or 90° - 55.89° = 34.11°

So the two possible launch angles for the object are approximately 55.89° and 34.11°. However, these values may change depending on the additional information about the object's mass and shape. It is also important to note that these are theoretical calculations and may not accurately represent real-world scenarios due to factors such as air resistance and the curvature of the Earth's surface.