1. Draw a picture I can't do that on here, so you will just have to do that part yourself. If you know anything about two dimensional motion and projectiles, you will know that you can find the velocity in the x and y directions with the information you are given: 56.6 m/s is equal to the initial velocity coming from the hose. To get the x and y velocities, we use the trigonometry associated with a right triangle and the angle of 42 degrees: SOH CAH TOA Sin Opp/Hyp Cos Adj/Hyp X is adj to the hyp, while y is opp the hyp. X uses cos and y uses sin. We know the hyp is 56.6 and we know that theta/angle is 42 cos(42) x/56.6 sin(42) y/56.6 x 56.6cos(42) 42.06199712 y 56.6sin(42) 37.87279232 Since we know the velocity for x, we can figure out how long it will take the water to travel the 38.7 feet. After finding that, we can put that time back into the y component and find the height. s v*t 38.7m 42.062*t t .92007 seconds Now that we have the time, we can use the y component to find the height: s ut + ?at? s (56.6sin42)(.92007) + ?(-9.8)(.92007)? s 34.84562003 + (-4.147991144) s 30.69762889 m
