Kinematics Min/Max angles of a projectile

[ieaglei]

Homework Statement

A football player attempts a 30m field goal. If he is able to impart a velocity u of 30 m/s to the ball,compute the minimum angle θ for which the ball will clear the crossbar of the goal. (Hint: Let m = tanθ .)

Homework Equations

V0x = V cosθ
V0y = V sinθ
We get t as a function of θ , using : X=X0 + V0xt => t=1/cosθ
And from the Y direction using Y = Y0 + V0yt + -g/2t^2
and then we substitute t from the x equation , in the y equation.

The Attempt at a Solution

Now , i substituted the both equation and after hours of trying and simplifying , i came up with this equation :
30sinθcosθ - (cosθ)^2 - 4.905 = 0
and i got stuck on how should we get 2 values of θ from this equation.
I really need help , i got the idea of the problem but i started giving up :(

 

[tiny-tim]

welcome to pf!

hi ieaglei! welcome to pf!

hmm …​

Hint: Let m = tanθ

 

[barryj]

You might consider this cheating but you could use a graphing calculator to solve.

Sure this is correct? - (cosθ)^2

 

[ieaglei]

I saw a solution for a very similar question , and they got an equation similar to my equation above , and then they wrote : Using trail and error we get : θ1 and θ2
and now I don't want to waste my whole life putting numbers in the calculator to find the angles , i want the right way to solve this :(
and I already tried plotting the graph , and what i got is a sinusoid with infinity solutions which made me more confused .
And by the way , i know the final answer which is θ mini = 15.43 degree

 

[barryj]

I don't think you entered the 3m crossbar heighth correctly in your equation. You used 1m.
I graphed the solution on my TI84 and got the correct answer.

 

[ieaglei]

hi ieaglei! welcome to pf!

hmm …​

Man , I saw a lot of problems on this site similar to mine , and I also saw your reply's :X
and i had no benefits of them , I want some serious hints , i want to go on and solve it :(.

 

[barryj]

First, get the correct equation to solve.
Then learn how to solve graphically if not analytically.

 

[ieaglei]

Oh , sorry the real equation i got was : 30sinθcosθ - 3(cosθ)^2 - 4.905 = 0
not : 30sinθcosθ - (cosθ)^2 - 4.905 = 0
Can you explain how you got the right answer?

 

[barryj]

I cheated and used a graphing calculator.
So far, I have not found an analytic solution.

 

[ieaglei]

I have the same TI-84 plus silver edition as a program on my pc , but I don't know how to use it , can you just tell me how to plot that equation :( , I know I'm asking so much .. but this is the last :)

 

[barryj]

Its easy. Plot y = 30sinθcosθ - 3(cosθ)^2 - 4.905 then look for the zero crossing. On a TI-84, once you see the graph cross zero, do (2nd)(trace)(2) to find the zero crossing. Put the cursor to the left of the crossing, then to the right of the corssing, hit enter and look at the answer.

 

[haruspex]

30sinθcosθ - 3(cosθ)^2 - 4.905 = 0

You were given the hint m = tanθ. Easiest is to divide through by cos²θ and then turn all the trig references into functions of m. you should get a quadratic.

 

[barryj]

Hmmmm... seems like if you divide by (cos theta)^2 you would have.
30 tan(theta) -3 -4.905/(cos(theta))^2 = 0

so, 1/(cos(theta)^2 = sec^2(theta)=1-tan^2(theta)

Ah Ha!

 

[haruspex]

Hmmmm... seems like if you divide by (cos theta)^2 you would have.
30 tan(theta) -3 -4.905/(cos(theta))^2 = 0

so, 1/(cos(theta)^2 = sec^2(theta)=1-tan^2(theta)

Ah Ha!

sec^2(theta)=1+tan^2(theta)

 

[ieaglei]

Thank you guys .. with your help , i reached the answer .
As you said : making the equation in terms of tanθ..
we will get : ( 4.905tan^2(θ) - 30tan(θ) + 7.905 = 0 )
replacing tanθ with x , we get a quadratic equation .. solving it we get 2 answers ..
tan^-1 (x1) = θ1
tan^-1 (x2) = θ2

problem solved and got the correct answer.
thread can be closed <3 thanks all

 

[mtram]

But i got theta is equal to 5.84 and 0.27 and when i subsitute in y equation to check my answer it gives me -1.8

 

[mtram]

oooh i got m..forget to take tan^-1...Thanks a lot for your help :)