• Support PF! Buy your school textbooks, materials and every day products Here!

A simple problem involving motion in two directions

  • #1
338
0
This is my first post, so an introduction is in order.

My educational background includes a MBA from UGA in 2007, a MS in EE from Southern Illinois U in 1994, and a BS in EE with minors in physics and math from SIU in 1992.

Recently I've applied to three universities for their PhD physics program. And I have to take the GRE Physics subject exam.

I've been studying using an old college physics text book by Ohanian. It's an old book, probably 20 years old or so. But I've been going through each chapter working the even numbered problems (because the answers are in the back). One particular problem has thrown me for a loop. Here it is...

You throw a stone at 25 m/s. Can you hit a window 50 m away that is 13 m above the ground? What is the highest the window can be at that distance that you can hit?
Here are some equations:

Vx = V cos(θ)
Vy = V sin(θ) - gt

x=Vt cos(θ)
y=Vt sin(θ) - ½gt²


I've been trying to solve this problem by writing y in terms of x and v, but I can't get rid of θ.
 

Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
3,566
189
This is my first post, so an introduction is in order.

My educational background includes a MBA from UGA in 2007, a MS in EE from Southern Illinois U in 1994, and a BS in EE with minors in physics and math from SIU in 1992.

Recently I've applied to three universities for their PhD physics program. And I have to take the GRE Physics subject exam.

I've been studying using an old college physics text book by Ohanian. It's an old book, probably 20 years old or so. But I've been going through each chapter working the even numbered problems (because the answers are in the back). One particular problem has thrown me for a loop. Here it is...



Here are some equations:

Vx = V cos(θ)
Vy = V sin(θ) - gt

x=Vt cos(θ)
y=Vt sin(θ) - ½gt²


I've been trying to solve this problem by writing y in terms of x and v, but I can't get rid of θ.
Just work with the last two equations (for x and y). You have two unknowns (t an dtheta) for two equations so it's solvable. To getrid of the angle, isolate sin and cos and then use that sin^2 + cos^2 = 1. That will leave you one equation containing x, y and t. You can therefore solve for t.
 
  • #3
338
0
Just work with the last two equations (for x and y). You have two unknowns (t an dtheta) for two equations so it's solvable. To getrid of the angle, isolate sin and cos and then use that sin^2 + cos^2 = 1. That will leave you one equation containing x, y and t. You can therefore solve for t.
Well, I get a mess that looks like this:


(½g²t²+yg-v²)t²+x²+y²=0
 
  • #4
nrqed
Science Advisor
Homework Helper
Gold Member
3,566
189
Well, I get a mess that looks like this:


(½g²t²+yg-v²)t²+x²+y²=0
you mean 1/4 for the first term.

Now plug in th evalues of x and y and g and solve for t.
 
  • #5
338
0
you mean 1/4 for the first term.

Now plug in th evalues of x and y and g and solve for t.
evalues? It's been almost 20 years since I've done stuff like this. Please clarify.
 
  • #6
nrqed
Science Advisor
Homework Helper
Gold Member
3,566
189
evalues?
I think that it's clear from my message that it's a typo, it should be "the values" instead of "th evalues"
It's been almost 20 years since I've done stuff like this. Please clarify.
You have an equation that contains t^4, t^2 and a cinstant. You can solve for the value of t^2 using the quadratic formula (in other words, just rename t^2=w, say. Then you have an equation of the form a+bw + cw^2 =0 with a,b,c being some constants. Use the quadratic formula to find w, which is t^2. )
 
  • #7
338
0
I see. I thought you might have been talking about eigenvalues.
 
  • #8
338
0
Need help solving this for θ:

a/cos²(θ)+btan(θ)-c=0

I've tried just about every trig identity known to man.

Thanks.
 
  • #9
338
0
a sec²θ + b tan θ - c = 0

use Pythagorean identity sec²θ = tan²θ + 1

a(tan²θ + 1) + b tan θ - c = 0

a tan²θ + b tan θ + (a - c) = 0

Use of the quadratic formula yields

tan θ = [-b ± √(b² - 4a(a - c))]/2a

θ = arctan{[-b ± √(b² - 4a(a - c))]/2a}
 

Related Threads for: A simple problem involving motion in two directions

Replies
6
Views
2K
Replies
6
Views
3K
Replies
11
Views
749
Replies
1
Views
2K
Replies
8
Views
2K
  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
17
Views
3K
Top