How to find angle whit system of equations

AI Thread Summary
To determine the angle to kick a ball with an initial velocity of 18 m/s to barely pass over a wall 8 m away and 1.5 m high, a system of equations is used. The vertical and horizontal components of velocity are expressed as 18sin(θ) and 18cos(θ), respectively. The key challenge is to eliminate time (t) from the equations, leading to a relationship involving both tangent and cosine of θ. By substituting the cosine squared identity, a quadratic equation for tan(θ) can be derived. Understanding vertical displacement (s_y) is crucial for solving the problem effectively.
felipenavarro
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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 can't soleve the equation. is this the right way to do it?
 
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So you have s_y = v_{oy}t -\frac{1}{2}gt^2, and you have values for s_y and v_{oy}.
If you know the wall is 8m away, and you know v_{ox}, then by rearranging s_x = v_{ox}t for t and subbing into the above eqn, you should be able to solve for θ.
It is just a case of rearranging the final eqn for θ. Is this the part you are having difficulty with?
 
sorry what is sy(changes in y?) , and yes that is where i am having trouble. i end up with cosines and sines and can't unite them or cancel them or anithyng
 
felipenavarro said:

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 can't 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:
sorry what is sy(changes in y?) , and yes that is where i am having trouble. i end up with cosines and sines and can't unite them or anithyng

s_y is the vertical displacement of the object. I implicitly assumed s_o = 0 in the equation s_y - s_o = v_{oy}t -\frac{1}{2}gt^2
 
now i got it, great explenation! thanks
 

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