I have submitted this question TWO TIMES and have gotten wrong answers both time

Question

I have submitted this question TWO TIMES and have gotten wrong answers both times. Thank you for your help. It is greatly appreciated

Problem 3.51 A firefighting crew uses a water cannon that shoots water at 28.0 m/s at a fixed angle of 46.0 o above the horizontal. The firefighters want to direct the water at a blaze that is 15.0 m above ground level. Part A How far from the building should they position their cannon? There are two possibilities (di d2); can you get them both? (Hint: Start with a sketch showing the trajectory of the water.) ANSWER: di Part B ANSWER:

Explanation / Answer

here,

initial speed of water , v = 28 m/s

theta = 46 degree

height of building , y = 15 m

let the distance of cannon from the building be x

using equation of trajectory for a projectile

y = x * tan(theta) - g * x^2 /( 2 * v^2 * cos^2(theta))

15 = x * tan(46) - 9.8 * x^2 /( 2 * 28^2 * cos^2(46))

solving for x

x = 19 m or 60.95 m

the two values of the distance from building are d1 = 19 m or d2 = 60.95 m

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