Lecture4-2A_000

Lecture4-2A_000 - Today's Lecture Motion in More Than One-D...

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Today Today s Lecture s Lecture Motion in More Than One Motion in More Than One - - D D Circular Motion 1 Circular Motion 1
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y tan x g 2 V 0 2 cos 2 x 2
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Example: Daredevil Motorcyclist Example: Daredevil Motorcyclist (a) What is the minimum speed for the cyclist to make the jump? This problem can be easily solved by making use of the parabolic trajectory equation. Solving for the initial velocity we find y tan x g 2 V 0 2 cos 2 x 2 with y 5.9 m , x 48 m , 15 2 V 0 2 cos 2 gx 2 tan x y V 0 1 2cos 2 gx 2 tan x y V 0 x cos g /2 tan x y 25.4 m / s 91 km / h
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Example: Daredevil Motorcyclist Example: Daredevil Motorcyclist (b) If he exceeds this velocity by 50% how far from the rim, d , will he land? Again, this problem is easily solved by making use of the parabolic trajectory equation. The resulting quadratic equation for
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Lecture4-2A_000 - Today's Lecture Motion in More Than One-D...

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