Okay I got it this time for sure :). Your right my 6.04 answer should have been 7.44 just a miscalculation. Once I worked it through I got 23.8 and 72.2 and those work in the equations :)
Thank you for your help
Homework Statement
Two boats are heading away from shore. Boat 1 heads due north at a speed of 3.43 m/s relative to the shore. Relative to Boat 1, Boat 2 is moving 39.6 ° north of east at a speed of 1.02 m/s. A passenger on Boat 2 walks due east across the deck at a speed of 1.58 m/s relative...
Oh the quadratic.
Ok so...
-26.5+√702.25-(4*-6.04*-8.85)
-------
2*-6.04
and the same but with a minus
-26.5-√702.25-(4*-6.04*-8.85)
-------
2*-6.04
so x=0.36 and 4.02
Then I plug it back in right? So I get:
tan-10.36=θ...
Okay. Using that this is what I'm getting:
2.81=26.5tanθ+(-6.04*(1+tan2θ))
So I should distribute?
2.81=26.5tanθ-6.04-6.04tan2θ
Put everything to one side?
0=-6.04tan2θ+26.5tanθ-8.85
What should I do now?
Okay that sound good in theory :). But my trig isn't that good so i get a little confused.
So far I have
26.5/(21.5cosθ)=t
Plugged it into get:
2.81=(21.5sinθ)(26.5/21.5cosθ)+(-4.9)(26.5/21.5cosθ)^2
Can I get some hint at what I should do next to be able to get rid of all the trig functions
So...
26.5=(21.5cosθ)t
and
2.81=(21.5sinθ)t+1/2(-9.8)t^2
is that the right next step? Solve the top equation for t and plug that t into the second equation to be able to solve for θ?
Homework Statement
A placekicker is about to kick a field goal. The ball is 26.5 m from the goalpost. The ball is kicked with an initial velocity of 21.5 m/s at an angle θ above the grojund. Between what two angles, θ1 and θ2, will the ball clear the 2.81-m-high crossbar? Give your answers as...