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

How to find angle whit system of equations

  • #1

Homework Statement


i kick a ball with an initial velocity of 18m/s. there is a wall at 8m that has a height of 1.5 meters. whith what angle should i kick the ball so that it bearly passes the wall?


Homework Equations


first equation: hf= 1/2gt^2+vit+hi
second equation: V=Δx/t

The Attempt at a Solution


so what i am trying to do is a system of equations isolating t in the second equation and placing it in the first equation. in velosity in the y direction i use 18sin(θ) and for v in the x i use 18cos(θ).
Also as final heigth i am using 1.6m(just a little higher than the wall) and as xfinal i am using 8.1m(also bearly passing the wall)

the problem is i cant soleve the equation. is this the right way to do it?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
CAF123
Gold Member
2,902
88
So you have [tex] s_y = v_{oy}t -\frac{1}{2}gt^2,[/tex] and you have values for [itex] s_y [/itex] and [itex]v_{oy}. [/itex]
If you know the wall is 8m away, and you know [itex]v_{ox}, [/itex] then by rearranging [itex] s_x = v_{ox}t [/itex] for [itex] t [/itex] and subbing into the above eqn, you should be able to solve for [itex] θ.[/itex]
It is just a case of rearranging the final eqn for [itex] θ. [/itex] Is this the part you are having difficulty with?
 
  • #3
sorry what is sy(changes in y?) , and yes that is where i am having trouble. i end up with cosines and sines and cant unite them or cancel them or anithyng
 
  • #4
ehild
Homework Helper
15,477
1,854

The Attempt at a Solution


so what i am trying to do is a system of equations isolating t in the second equation and placing it in the first equation. in velosity in the y direction i use 18sin(θ) and for v in the x i use 18cos(θ).
Also as final heigth i am using 1.6m(just a little higher than the wall) and as xfinal i am using 8.1m(also bearly passing the wall)

the problem is i cant soleve the equation. is this the right way to do it?
It is the right way in principle but "just a little higher" can mean anything. Why not 1.55 m? or 1.51 m? So choose y=1.5 m at x=8 m.
After eliminating t, you get an equation which contains both the tangent and the cosine of theta. (y=tan(θ)x-(g/2)x2/(182cos2(θ)). You certainly know that cos2(θ)=1/(1+tan2(θ)). Use that relation and you get a quadratic equation for tan(θ).

ehild
 
Last edited:
  • #5
CAF123
Gold Member
2,902
88
sorry what is sy(changes in y?) , and yes that is where i am having trouble. i end up with cosines and sines and cant unite them or anithyng
[itex] s_y [/itex] is the vertical displacement of the object. I implicitly assumed [itex] s_o = 0 [/itex] in the equation [tex] s_y - s_o = v_{oy}t -\frac{1}{2}gt^2 [/tex]
 
  • #6
now i got it, great explenation! thanks
 

Related Threads on How to find angle whit system of equations

  • Last Post
Replies
1
Views
911
Replies
3
Views
7K
Replies
5
Views
167
Replies
0
Views
1K
  • Last Post
Replies
6
Views
15K
Replies
7
Views
2K
Replies
3
Views
11K
Top