SOLUTION: Water shoots out from a fountain at 40 m/s at an angle of 55° above the horizontal. What is its time of flight? What is its range?

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Question 1176901: Water shoots out from a fountain at 40 m/s at an angle of 55° above the horizontal.
What is its time of flight?
What is its range?
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Water shoots out from a fountain at 40 m/s at an angle of 55° above the horizontal.
What is its time of flight?
vel= 40m/s
angle =55
V up = 40 sin (55)= 8.2 m/s
v= u +at
t=(v-u)/g
t=(0-8.2)/-9.8
t= 0.84 s

Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.


            The solution / (the calculations)  in the post by   @mananth are     INCORRECT.

            I came to bring you a correct solution. 


Vertical component of the initial velocity is  V%5Bvert%5D = 40*sin(55°) = 40*0.8189 = 32.75 m/s.



The time for water to get the maximum height is  T%5Bmax_height%5D = V%5Bvert%5D%2Fg = 32.75%2F10 = 3.28 seconds.



The time of the flight is  (assuming that the hose is at the ground level)


    T%5Bflight%5D = 2%2AT%5Bmax_height%5D = 2*3.28 = 6.55 seconds.



The maximum height = %28g%2AT%5Bmax_height%5D%5E2%29%2F2 = %2810%2A3.28%5E2%29%2F2 = 53.8 meters.

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What is  "the range"  in your last question is unclear to me,  so I stop my solution at this point.