What Angles Allow a Golf Ball to Land 85 Meters Away?

[physics_noob2]

Homework Statement

Physics question: A golfball with an initial speed of 57.9 m/s lands exactly 85 m downrange on a level course. The acceleration of gravity is 9.8 m/s^2
Neglecting air friction, what minimum projection would achieve this result ? What maximum projection would achieve this result ? Answer in units of degrees Please.

Homework Equations

The Attempt at a Solution

Viy=57.9 sin theta
ay= -9.8
delta y=0

vix=57.9 cos theta
ax= 0
delta x=85

from horizontal,
t= 85/(57.9 cos theta)
from vertical,
0 = (57.9 sin theta) t + 1/2 (-9.8) (t)^2


So i tried doing it and got to the point so far and got stuck (can't seem to do the algebra). can anyone show me how to do this please Thanks.

 

[Delphi51]

Sub the expression you have for t into the vertical equation. Get all your trig functions together on the left and the numbers on the right. You'll have sin(θ)*cos(θ) = a number.
Use the trig identity that sin(θ)*cos(θ) = ½*sin(2θ)
and you'll quickly get one value for 2θ. The other angle that gives the same sine is 180 degrees minus the first value (check it out on a unit circle sketch). Divide by 2 to get the two values for θ.

It is just a little easier if you use V = Vi + at for the vertical part instead of the distance formula. I always begin by writing both because you never know in advance which will be easier to work with.