Answer:
[tex]x=1.071m[/tex]
Explanation:
Given data
[tex]v_{i}=1.5m/s\\v_{f}=0\\ y_{i}=0m\\x_{i}=0m\\y_{f}=-2.5m[/tex]
To find how far out from the end of the pipe is the point where the stream of water meets the creek
From equation of simple motion
[tex]y_{f}-y_{i}=+v_{i}t+1/2at^{2} \\y_{f}=y_{i}+v_{i}t+1/2at^{2} \\-2.5m=0+0+(1/2)(-9.8m/s^{2} )t^{2}\\ 2.5m=(1/2)(9.8m/s^{2} )t^{2}\\t^{2}=\frac{2.5}{4.9} \\ t=\sqrt{\frac{2.5}{4.9}}\\ t=0.714s[/tex]
As we know ax=0 m/s² for projectile
We can find required distance x easily
[tex]x=v_{i}t\\x=(1.5m/s)(0.714s)\\x=1.071m[/tex]
