Check if Solution is correct.

  • Thread starter polak333
  • Start date
  • #1
24
0

Homework Statement


Derive an equation for the horizontal range of a projectile with a landing point at a different altitude from its launch point. Write the equation in terms of the initial velocity, the acceleration due to gravity, the launch angle, and the vertical component of the displacement. Please check to see if it is correct!
I tried to make the formulas as realistic as possible.

p.s. the "/" tend's to mean a fraction. a/b = a/b

Homework Equations



x = vcosΘt, t = v/cosΘ (1)

y = vsinΘt - 1/2 gt2 (2)

The Attempt at a Solution



Solving (1) for t and substituting this expression in (2) gives:

y = xtanΘ - (gx2/2v2cos2Θ)

y = xtanΘ - (gx2sec2Θ/2v2) .....(Trig. identity)

y = xtanΘ - (gx2/2v2) * (1+tan2Θ) .....(Trig. identity)

0 = (-gx2/2v2 * tan2Θ) + xtanΘ - (gx2/2v2) - y .....(Algebra)

Let p = tanΘ​

0 = (-gx2/2v2 * p2) + xp - (gx2/2v2) - y .....(Substitution)

p = -x ± √{ x2 - 4(-gx2/2v2)(-gx2/2v2 -y)} .....(Quadratic formula)
,,,,,,,,,,,,,,,,,,,2 (-gx2/2v2)

p = v2 ± √{ v4 - g(gx2 + 2yv2)} .....(Algebra)
,,,,,,,,,,,,,,,,,,,,,,,,,gx

tanΘ = v2 ± √{ v4 - g(gx2 + 2yv2)} .....(Substitution)
,,,,,,,,,,,,,,,,,,,,,,,,,gx

Θ = tan-1 [v2 ± √{ v4 - g(gx2 + 2yv2)}]
,,,,,,,,,,,,,,,,,,,,,,,,,gx
 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
11
Your algebra all looks good! But it seems to me you are asked to find x as a function of the initial speed, angle and y. That would mean starting with x = vt*cos(theta) and trying to solve the y equation for t in order to eliminate t in the expression for x.
 
  • #3
24
0
Hmm. I see, I guess I just started solving for anything else.
Do you have any suggestions how I can do this? or how it looks like?
 
  • #4
38
0
Let the desired horizontal range be h while the corresponding vertical height is k. We have:

[tex]h= v_{0}\cos(\theta)t \iff t = \frac{h}{v_{0}\cos(\theta)} \, * [/tex]

And

[tex]k = v_{0}\sin(\theta)t + \frac{1}{2} g t^{2}[/tex]

Use the substitution given by (*) to get:

[tex] k = v_{0}\sin(\theta) \left( \frac{h}{v_{0}\cos(\theta)} \right) + \frac{1}{2} g \left( \frac{h}{v_{0}\cos(\theta)} \right)^{2} [/tex]

[tex]k = \tan(\theta)h + \frac{gh^{2}\sec^{2}(\theta)}{2v^{2}_{0}}[/tex]

This is as far as you need to go since you have a relationship in terms of all the variables asked for in the question. It might be possible to isolate h in the equation, though I have no idea how it would be done.
 
  • #5
24
0
Whew.. all that work for nothing.
Thanks mate!
 

Related Threads on Check if Solution is correct.

  • Last Post
Replies
1
Views
1K
Replies
3
Views
2K
Replies
2
Views
650
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
2
Views
2K
Replies
12
Views
2K
Replies
1
Views
1K
Replies
5
Views
648
Top