posted by maddi shane .
I'm having a hard time answering this question. I don't know what equation to use? I've tried using dy= Viyt + 1/2 gt^2. Water is leaving a hose at 6.8 m/s. If the target is 2 m away horizontally, What angle should the water have initially? thank you :D
Range = (Vo^2/g)*sin(2A) = 2.0 m
Vo = 6.8 m/s is the initial velocity.
Solve for A.
sin(2A) = 0.4239
2A = 25.08 or 154.92 degrees
A = 12.54 or 77.46 degrees
Thank YOU :D
Aunt "Maddy Shane" was the nickname given to my grandmother by her nieces. Shane was the maiden name.
I never thought I'd see that name again.