Posted by **maddi shane** on Thursday, November 24, 2011 at 5:22am.

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

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**drwls**, Thursday, November 24, 2011 at 5:57am
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

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**maddi shane**, Thursday, November 24, 2011 at 6:40am
Thank YOU :D

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**drwls**, Thursday, November 24, 2011 at 2:23pm
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.

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