Maxiumum height and range

In summary, the angle of projection for a projectile fired in a way that its horizontal range is equal to three times its maximum height is about 53 degrees. This can be calculated using equations for range and height, as well as energy conservation, and taking a ratio of the initial vertical and horizontal velocity components. Some basic trigonometry is also necessary to solve for the angle.
  • #1
niyati
63
0
A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection?

If I let A = maximum height (or h), than horizontal range (or R) = 3A.

I suppose I can also substitute 1 for h and 3 for R. There is some relationship between the horizontal velocity and the initial vertical velocity that allows for this feat to be accomplished. The angle must be of about medium amount, about 40-60 degrees?

Equations include the components of vf = vi + at, components of rf = ri + vit + .5a(t^2), equations for range and height:

h = ((vi^2)((sinθ)^2))/(2g), where g = -9.8 m/s^2
time it takes for particle to reach max height: th = (vi(sinθ))/g
R = vxi2th = ((vi^2)(sin2θ))/g

θ = angle of projection, but all of these equations require some knowledge of the initial velocity, which I don't have. And the equations involving that include time, which I also do not have. I think I'm not getting something important from the information already given. The point at the beginning is (0,0), while at maximum height, the coordinates are (1.5,1) or (3A, A).

So, now I think that I can use the rules of parabolic equations to figure this out. But, I don't think they want me to figure this problem out like that.
 
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  • #2
What you need to find are the ratio of horizontal and vertical velocities at the start.
For the horizontal component there is no accelaration so s(h) = v(h) * t
Vertically you have the famous s = ut + 1/2 at^2 use this for the vertical motion. Remember the time to reach the top of the parabola is half the time to complete the flight.
You should be able to end up with values for v(horiz) and v(vert) then a simple bit of trig will give the angle.
 
  • #3
But if I do not know the time, nor the value of the initial velocity vectors, how can I figure it out?
 
  • #4
You are going to get equations for the initial vertical and horizontal velocity.
They might have 't' in them but you are going to take a ratio of the two velocity components to get the overall direction so these will cancel out.

Or you can use energy conservtion:
Vertically at the start it has only ke = 1/2 m v(v)^2
At the top it has only potential pe = m g h
Mass cancels so;
g R = 0.5 * v(v)^2 or V(v) = sqrt(2 g R )

You can then use s = ut + 1/2 at^2 to get the time.
Then use this time to get the horizontal speed, remember that 't' is the time to get to the top of the arc so is half the time to travel the horizontal distance.
 
  • #5
I am still having trouble with this problem, although I sincerely appreciate your help.

For x: 1.5 = vxit or t = 1.5/vxi

As for y: 1 = vyit - 4.9(t^2)

I understand that while x and y are of independent paths, they both have something in common, usually either a time or an angle. In this case, it is time. I do not think I have to alter the variable of time in any way, as the final x and y points I am dealing with are the ones at the pinnacle of the parabola. Therefore, the time should only be of half the trajectory. Correct me if I am wrong in following this.

I've converted the x part into t = 1.5/vxi, and tried plugging it into what I had for y, and I don't think that's right either and, well, yea.

X/
 
  • #6
You've already done most of the work in your first post. You've calculated the height in terms of v and [tex]\theta[/tex]. You've calculated the range in terms of v and [tex]\theta[/tex]

divide the h equation by the R equation... v^2 cancels out. h cancels out since R = 3h. Now you can get [tex]\theta[/tex]
 
  • #7
Ooooookay, so:

1/3 = (sinθ)^2/(2sin2θ)

Therefore:

(3/2) = sin2θ/(sinθ)^2

But, that requires sin2θ to equal three, and that is not possible.
 
  • #8
niyati said:
Ooooookay, so:

1/3 = (sinθ)^2/(2sin2θ)

Therefore:

(3/2) = sin2θ/(sinθ)^2

But, that requires sin2θ to equal three, and that is not possible.

expand sin2θ then simplify.
 
  • #9
OH crudmuffins!

I completely forgot about...well, the 23 basic trig identities I had (note the past tense) memorized about a year ago.

All right, all right. It's about 37 degrees. Thank you...so much for you patience and help. :D I appreciate it greatly.
 
  • #10
niyati said:
OH crudmuffins!

I completely forgot about...well, the 23 basic trig identities I had (note the past tense) memorized about a year ago.

All right, all right. It's about 37 degrees. Thank you...so much for you patience and help. :D I appreciate it greatly.

It's not 37 degrees, you're close through... you made a small mistake. you mixed up 3/4 and 4/3... :wink:

The answer's 53 degrees.
 
Last edited:
  • #11
...

>_<

Well, this is embarrassing.

XD
 
  • #12
niyati said:
...

>_<

Well, this is embarrassing.

XD

Don't be embarrased. I've lost count of how many times that type of thing has happened to me! :smile:
 
  • #13
M'kay. I actually got that answer you did.

I took the tangent instead of the cotangent, the way I divided over all the numbers.
 
  • #14
niyati said:
M'kay. I actually got that answer you did.

I took the tangent instead of the cotangent, the way I divided over all the numbers.

Cool. :cool:
 

What is maximum height and range?

Maximum height and range refer to the furthest distance an object can travel and the highest point it can reach, respectively. It is commonly used in physics and engineering to understand the capabilities and limitations of various objects.

How is maximum height and range calculated?

The maximum height and range of an object can be calculated using various equations, depending on the initial velocity, angle of launch, and other factors. The most commonly used equation is the projectile motion equation: h = (v2 * sin2θ) / 2g for maximum height and R = (v2 * sin 2θ) / g for maximum range.

What affects the maximum height and range of an object?

The maximum height and range of an object can be affected by several factors, including initial velocity, angle of launch, air resistance, and gravity. The shape and weight of the object can also play a role in determining its maximum height and range.

Why is understanding maximum height and range important?

Understanding maximum height and range is essential for various fields of study, such as physics, engineering, and aviation. It helps in designing and testing objects, predicting their capabilities and limitations, and ensuring their safe use.

Can maximum height and range be improved?

Yes, maximum height and range can be improved by optimizing various factors, such as the shape, weight, and initial velocity of the object. Additionally, reducing air resistance and using advanced propulsion systems can also increase an object's maximum height and range.

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