Recent content by theoronhalde

oh...i'm so lost once again. I tought we should get the time By using: y= yo+Voy*t-(1/2)gt^2 formula, = 3.25+ (13.3sin75)t-(1/2)(-9.81)t^2 If not can you explain me how to find the time and how to get the components Vfy and Vfx.

oh ok...i undestand what you are saying but i still end up with a wrong answer...here is what i did: Vfx= Vox + at (a=0) =13.3cos75 Vfy= Voy+at = 13.3sin75 + (9.81)(2.85) = -15.1 Vf= SQRT( Vfy^2 + Vfx^2) = 15.5 m/s The applet says that the answer is wrong :S ??

ok s it's going to be: Vfy= Voy+at = 13.3sin75+ (-9.81)(2.85) = -15.1 m/s But how come the horizontal velocity be constant? And am I suppose to use V=Sqrt((Vox^2)+(Vfy^2)) to find that final speed?

hi! thanks a lot! Not sure i understand your answer? Do i have to use the formula you wrote instead of the one i was using to fnd the final velocty? I'm a bit confuse...

final velocity...need help! Homework Statement An object is to be thrown through a window 3.25 m high a distance 8 m away. It is launched at an angle 75 degree above the horizontal. The initial launch speed at the specified angle which will allow the projectile to pass through the window is...