Using constant acceleration (acc'n) equations: i=horizontal componant, j=vertical
(displacement) S={Uit+0.5ait²} + {Ujt+0.5ajt²}
(Initial Velocity) U={21.9*cos(36.3)i} + {21.9*sin(36.3)j}
(Final Velocity) V={Ui+ait} + {Uj+ajt)
(Acc'n) a={0i} + {-9.81j}
(time) t=t
When the water is at it's highest point, Vj=0, since no vertical motion is occurring
therefore:
21.9*sin(36.3) + -9.81 = 0
therefore, when the water is at it's highest point, t=1.32 (3 sig figures)
To find how far away you need to be, you need to work out how far the water would travel in this time, using Si
Therefore:
when t=1.32.....
Si= {21.9*sin(36.3)}*(1.32....) + 0.5*9.81*1.32.... = 25.66
