Finding an angle of a projectile for a specific distance

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SUMMARY

The discussion focuses on calculating the launch angle of a fire hose shooting water at a speed of 6.5 m/s to achieve a horizontal distance of 2.0 meters. The relevant equations include the kinematic equations for projectile motion, specifically separating the vertical and horizontal components of velocity: Vy = Vosinx and Vx = Vocosx. The solution involves treating the x and y directions independently, applying gravitational acceleration in the vertical direction, and ultimately deriving two possible angles due to the nature of projectile motion.

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trexmatt
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Homework Statement



A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the water land 2.0m away. Why are there two different angles?

Homework Equations



v=v0 +at v^2=v0^2 +2ad d=V0t+1/2at^2

The Attempt at a Solution



I'm not sure how to go about solving this. I've gotten as far as this (calling the angle x)

Vy = Vosinx and Vx = Vocosx

Any help would be much appreciated, I don't understand where to go from here. It seems like none of the equations are going to go anywhere.

Thank you very much for your time...
 
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Welcome to PF!

Hi trexmatt ! Welcome to PF! :smile:

(have a theta: θ and try using the X2 tag just above the Reply box :wink:)
trexmatt said:
v=v0 +at v^2=v0^2 +2ad d=V0t+1/2at^2
…
Vy = Vosinx and Vx = Vocosx

Yes, those are the equations to use :smile:

(except don't use x for two different things! :rolleyes:)

Now treat the x and y directions separately (a = -g for y, a = 0 for x), to get a pair of equations involving θ, from which you can then eliminate t.
 

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