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

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Question 1191924: Water shoots out from a fountain at 40 m/s at an angle 55 ° above the horizontal.
A. What is the time of flight
B. What is the range?
Answer by ikleyn(52814) About Me  (Show Source):
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Water shoots out from a fountain at 40 m/s at an angle 55 ° above the horizontal.
A. What is the time of flight
B. What is the range?
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(A)  Calculate verical component of initial speed 

         V = 40*sin(55°) = 40*0.8192 = 32.768 m/s.


     Next, calculate the time moving up

         t = V%2Fg = 32.768%2F9.81 = 3.34 seconds  (where g= 9.81 m/s^2 is the gravity acceleration).


     Finally, time of flight (up and down)  is  T = 2*t = 2*3.34 = 6.68 seconds.    ANSWER



(B)  The range is the horizontal distance

         range = 40*cos(55°)*T = 40*0.5736*6.68 = 153.27 meters  (rounded)   ANSWER

Solved.