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.
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
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