Please show work (on paper if possible). Thank you for helping me! (1) A dop of

Question

Please show work (on paper if possible). Thank you for helping me!

(1) A dop of water falls with no intitial speed from poin A of a highway overpass. After dropping 6m, it strikes the windshield at point B of a car which is traveling at a speed of 100 km/h on the horizontal road. If the windshield is inclined 50deg from the vertical as shown, determine the angle (theta) relative to the normal n to the windshield at which the water drop strikes.

FYI the answer is (theta) = 28.7 deg below normal ... However I dont know how to get there/show my work.

Representative Problems 1 A drop of water falls with no initial speed from point A of a highway overpass. After dropping 6 m, it strikes the windshield at point B of a car which is traveling at a speed of 100 km/h on the horizon- tal road. If the windshield is inclined 50° from the vertical as shown, determine the angle relative to the normal n to the windshield at which the water drop strikes. 6 m rt 50° 100 km/h ! Problem 2/191

Explanation / Answer

After dropping the speed of the rain drop with respect to ground = Vrg =sqrt(2gh)

= sqrt(2*9.8*6)

= 10.8 m/s vertically downwards

= -10.8 j m/s <---------- in vector notation

Now, speed of the windsheild with respect to ground , Vwg = (100 km/h) i <----- in vector notation

= 100000/3600

= 27.8 m/s i

So, velocity of windsheild with respect to rain, Vwr = Vwg - Vrg

= 27.8 i + 10.8 j

So, the angle it makes with the horizontal = atan(10.8/27.8) = 21.2 deg

The normal makes 50 deg with the horizontal

So, the angle it makes with the rain = 50 - 21.2 = 28.8 deg below normal <---------answer

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