Projectile Angle: Finding the Launch Angle

[airforce840]

Suppose that a projectile is launched from an initial height yi with an initial speed Vi and it strikes a target at a fianl height yf and fianl position xf. Show that the proper launch angle (feta symbol) is given by

feta = tan^-1 [xf +/- Square root of Xf^2 - 4k(k-Delta y) / 2K]

where K = ayXf^2/ 2 Vi^2

sry for the HORRIBLE typing of formulas..i don't know how to add them in.

and i don't have a photocopier.

thanks in advanced for any help on this one.

Patrick

 

[finchie_88]

I'm assuming that air resistance is not being taken into account here.
So, initial velocity can be split into x and y components, from that, the position as a function of time can be obtained, and then solve for the angle.
(A is the angle of projection)
x-component:
$$v_x = v_0cosA$$
y-component:
$$v_y = v_0sinA$$
Therefore, the postions are:
$$s_x = v_0tcosA<br /> s_y = v_0tsinA - \frac{gt^2}{2}$$
Then re-arrange the first one of these for t, then substitute into the second equation, and solve for A (you should get a quadratic in tanA). You will need to take into account the initial height at the time of projection, but it should work. By the way, sorry about the dodge formulas, I'm not very good at using them, suppose that its lack of practise...

 

[airforce840]

uh... is there sumthing else that means S because I've never seen Sx before.. lol


other than that i kinda understand and yes..windresistance is always ignored in our class.. i don't know why but our teacher will not go into it

 

[airforce840]

i figured out most of it.. but i confused my self on this because there's 2 ways to do it. the 1st equation to t and then sx.. and u get this

t= -vi sin(feta) +- Square root of vi sin (feta) - 4(-sy) / 2

and if i use the other one.. its more complicated i think sx= vi(-vi Sin(feta)+-square root of (vi Sin (feta) - (4)(/-sy) / 2) cos (feta)

how do i solve for feta with EITHER of these horribly confusing mess' and if someone can tell me how to write these equations with those images..lol that would be nice..cuz i think it makes it look worse.
thanks in advanced

Patrick

 

[Gamma]

Read carefully what finchie_88 said. He has said every thing you need to do to get tan (A) or launch angle.

There is one way to solve this and there are two equations with two unknowns namely tan A and time 't'. You need to eliminate t first. In order to do that, find 't' from the first equation.

$X_f = v_i cos(A) t$

and substitute this one in the second equation.

$\Delta y = -v_i sin(A) t + \frac{gt^2}{2}$

Do some re-arranging and you will end up with the following quadratic equation,

$\frac{g X_f ^2}{2v_i ^2} tan^2 A - X_f tan A + (\frac{g X_f ^2}{2v_i ^2} - \Delta y) = 0$


i.e.

$K tan^2 A - X_f tan A + (K - \Delta y) = 0$

I am sure you can solve this.

Basically what we have done is using s = ut + 1/2 g t^2 horizontally and vertically to the given problem and the rest of it is just mathematics.

Hope this helps.