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

Kinematics in Two Dimensions

  • Thread starter kaspis245
  • Start date
  • #1
189
1

Homework Statement


I need to find velocity v from the drawing.

Homework Equations


Equations of kinematics

The Attempt at a Solution



I found that v=12.134 m/s . I need to know is it correct.
[/B]
image.jpg
 

Answers and Replies

  • #2
Delta2
Homework Helper
Insights Author
Gold Member
2,955
1,012
Your mistake should be in the calculation of s2. Since you use the equation [itex]h=1/2gt_{2}^2[/itex] seems like you ignore the fact that the particle has vertical velocity [itex]v_{y}=gt_{1}/2[/itex] at that point. So it should be [itex]h=v_{y}t_{2}+1/2gt_{2}^2[/itex]
 
  • Like
Likes kaspis245
  • #3
907
88
I get a different answer.

Maybe you can verbally explain your reasoning?
 
  • #4
I get a different answer.

Maybe you can verbally explain your reasoning?
Delta means this:in order to find your time for s2, you find the time it takes to fall a distance h under gravity with an initial velocity (in the y direction) of zero. This will give you the wrong time. You need to account for the initial vertical velocity. Think: which takes longer to hit the ground: a ball shot straight forward or a ball shot upward at an initial positive angle above the horizontal?
 
  • #5
907
88
I think it's a mistake to combine s1 and s2 the way you are doing because they depend on t^2.

The standard way to solve a problem like this is to break into two parts: Vx and Vy and use t as the common variable.
 
  • #6
189
1
I understand my mistake, it should be h=vyt2+1/2gt22

Here's what I get:

image.jpg


Then I tried to solve it:

image.jpg


I think this answer is wrong. Please help!

By the way, I did solve this equation using Wolfram Alfa and got v=6,88 m/s.
 
Last edited:
  • #7
Delta2
Homework Helper
Insights Author
Gold Member
2,955
1,012
You got drown in a spoon of water :D, you forgot the square root in the last step, the velocity is the square root of 47.4.

But i am not quite sure the initial equation ([itex]sg=2v_x^2+...[/itex] is correct can u show us how u derive it.

Ok i think it must be correct cause solving the problem with the "usual" aaproach described briefly by paisiello2 gives the same result.
 
Last edited:
  • Like
Likes kaspis245
  • #8
ehild
Homework Helper
15,492
1,874
There is a much simpler method based on the trajectory of the projectile. The horizontal displacement (x) is x = Vo cos(α) t , the vertical displacement y - h= Vosin(α) t -g/2 t2. Eliminate the time t, you get the equation y(x) for the whole trajectory:
## y = h + \tan(α) x - \frac{g x^2}{2 V_0^2 \cos^2(α)} ##
You know both the initial point and the final point of the trajectory: if x=0, y=h=0.17 m and x=s=5 m, when y=0. Alpha is given, α=45°. Substitute the known data, and solve the equation for Vo.
 
Last edited:
  • Like
Likes kaspis245

Related Threads on Kinematics in Two Dimensions

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
4
Views
15K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
815
  • Last Post
2
Replies
33
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
8
Views
2K
Top