blocked page by Websense, getting around it
Carbon monoxide. Tell her to get out of the house at once.
guess you're going to fail physics, won't be the first or the last. think gravity. go from there.
This is all I can do. Try harder.
Are you allowed to derive the Range Equation? (Easy way out) 1) Horiz Range: x(t) Vo* cos (Launch angle) * t(From here out symbol for launch angle () 2) Vertical Range y(t) Vo * sin ()*t - 1/2 gt^2 We need the time when y 0 (Launch and when the projectile hits the ground) t0 or Multiple 2 through by 2 to clear the 1/2 Y(t) 0 2*0 2Vo*sin()*t - g*t^2 Factor a t 0 t(2Vo*sin () - g*t) t0 or 2Vo*sin() - g*t 0 - g*t - (2Vosin() ) t (2Vosin() )/g Substitute t back in 1 x(t) (Vo * cos() ) * ((2Vo*sin () ) / g) 3) x(t) (2*Vo^2 cos() * sin() )/g Trig Identity sin (x+y) sin (x) * cos (y) + sin(y)*cos(x) But xy launch angle sin (2*launch angle) 2 (sin() * cos()) Substitute back into 3 x (Vo^2/g) * sin(2*launch angle) And we have the range equation Can you solve it from here? EDIT: Or Plug your numbers into the top and work the derivation to the bottom, putting in your numbers. The solution will be the minimum angle. Starting from horizontal, what is the first angle that works? If the angle is less than 45 degrees, there is a second angle 2 theta in the 2nd quadrant that will also work. Example: 2 Theta 60 degrees (sin 2 theta 0.866) Theta 30 degrees 2nd quadrant solution 180 - 60 120 degrees (sin 2 Thera sin 120 0.866) Theta 60 degrees Either 30 degrees or 60 degrees will work
