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The aim of this task is to use a hose, shooting out water at 10ms-1 and watch the changes in the flow of water at different positioning angles of the hose.


From this we will be able to obtain some knowledge of the graphs the water jet forms and its equations.


We will mathematically observe the actions of the water, but certain assumptions are made.


If the water jet was shot in real life there would be outside factors involved, such as


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Length of hose might decrease the speed of the jet


Jet not shooting straight and/or shooting some water to the side so the measurements are inaccurate


Wind influencing the movement of the water stream.


Different gravitational pulls at different heights.


In this problem-solving task these things are ignored and perfect conditions are assumed.


For increased accuracy (not perfect) in calculations working out will be shown to 5 decimal places.


In all cases when x=0 one of the solutions for y will be 0


The graphs that are sketched are drawn separately as this way is neater and easier to observe.


When we do shoot a jet out of a hose we notice that it forms a parabolic graph, so we can safely assume that jets at all angles will provide the same results, except when the stream is being shot at 0¢ª.


The equation of the graph that the jet forms is given by the function y=ux-0.04(1+u)x where u=Tan¥è


The number 0.04 is related to the speed of the water, 10ms and the acceleration due to gravity, .8ms-


For the case where ¥è=0¢ª, for us to be able to use the equation we need to find the Tan 0¢ª which is 0.5775


Therefore u=0.5775


When this is substituted into the equation we get a graph that looks like this


GRAPH


To find the place where the jet hits the ground we solve the equation so that y=0


When y=0, ux-0.04(1+u)x=0


x(u-0.041.x)=0


Therefore x=0 and x=8.8744


Since this is a parabolic graph the maximum point is the middle point for the x values and in this case that is 8.8744/ = 4.41850 and when x=4.41850, y=1.7551


Therefore the jet reaches the maximum height reached for the angle of 0¢ª is 4.41850m


For the case where ¥è=45¢ª, tan45¢ª=1


Therefore u=1


When this is substituted into the equation we get a graph that looks like this


GRAPH


Once again to find the spot where the water hits the ground we solve the equation for y=0


When y=0, x=0 and x=10.0408


The highest point is once again found by taking the half way point on the x-axis and finding the f(x).


Therefore for the ¥è=45¢ª, the maximum is .55000


For the angle ¥è=60¢ª, u=tan¥è therefore u=


When this is substituted into the equation we get a graph that looks like this





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