Engineering Mechanics, Dynamics
Favorites|Homepage
Subscriptions | sitemap
HOME > Physics > Engineering Mechanics, Dynamics

Engineering Mechanics, Dynamics

[From: ] [author: ] [Date: 11-09-26] [Hit: ]
is is 90º - 33.8º = 56.......
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
horizontal road. If the windshield is inclined 50 from the vertical as
shown, determine the angle θ relative to the normal n to the
windshield at which the water drop strikes.

-
The velocity of the drop after falling h m is v = √[2*g*h]

for h = 6 m and g = 9.8 m/s² v = √2*9.8*6] = 10.8 m/s

The angle of impact with respect to the vertical is arctan(vx/vy) = arctan(100/10.8) = 83.8º; since the windshield is 50º from vertical, the impact angle is 33.8º with respect to the windshield. With respect to the normal, is is 90º - 33.8º = 56.2º
1
keywords: Dynamics,Engineering,Mechanics,Engineering Mechanics, Dynamics
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .