Water is leaving a hose at 6.8 m/s.  If the target is 2 m away horizontally, what angle should the water leave the hose at? Answer is supposed to be 13 degrees but I am having trouble getting that.  Please help!

Expert Answers

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Let us say the angle to ground is theta;

Using equation of motion;

`rarr S = ut+1/2at^2`

At horizontal direction a = 0

`2 = 6.8costhetaxxt ----(1)`


`uarr S = ut+1/2at^2`

There is no vertical distance for the whole motion.

`0 = 6.8sinthetaxxt-1/2xx9.81xxt^2`

`2xx6.8sintheta = 9.81t ----(2)`



`1/(6.8sintheta) = 6.8costheta/9.81`

`9.81/6.81^2 = sinthetacostheta`


We know that;

`sin(2theta) = 2sinthetacostheta`


`9.81/6.81^2 = sinthetacostheta`

`9.81/6.81^2 = (sin(2theta))/2`

`sin(2theta) = 0.423`

`2theta = sin(-1)(0.423)`

`theta = (sin(-1)(0.423))/2`

`theta = 12.5 deg. ` (for small angles)


So the angle is 12.5 degrees to the ground.

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