# 2-D and projectile motion Answers - Physics 125...

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Physics 125 Discussion #4 Solution 1 Problem 1: A body projected horizontally A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0 m/s. x direction y direction initial position x 0 = 0 y 0 = 0 initial velocity v 0x = +9.0 m/s v 0y = 0 acceleration a x = 0 a y = - 9.8 m/s 2 (3) What is the motorcycle’s position ( x , y ) at t = 0.50 s? (m) 2 . 1 ) 50 . 0 )( 8 . 9 ( 2 1 0 2 1 (m) 5 . 4 ) 50 . 0 )( 0 . 9 ( 2 2 0 0 - = - + = + = = = = at t v y t v x y x (1) First, let’s define our coordinate system. The coordinate system is shown in the figure. We choose the origin at the edge of the cliff, where the motorcycle first becomes a projectile. (2) List all the known factors (in both x- and y- directions). 0 v h x y O

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Discussion #4 Solution 2 (4) What is the motorcycle’s distance from the origin at t = 0.50 s? (m) 7 . 4 ) 2 . 1 ( ) 5 . 4 ( 2 2 2 2 = - + = + = y x r (5) What are the x- and y- components of velocity at t = 0.50 s? (m/s) 9 . 4 ) 50 . 0 )( 8 . 9 ( 0 (m/s) 0 . 9 0 0 - = - + = + = + = = at v v v v y y x x The motorcycle has the same horizontal velocity v x as when it left the cliff at t=0, but in addition there is a vertical velocity v y directed downward (in the negative y-direction. (6) Express the velocity at t = 0.50 s in terms of magnitude and direction. o x y y x v v v v v 29 0 . 9 9 . 4 tan tan (m/s) 2 . 10 ) 9 . 4 ( ) 0 . 9 ( 1 - 1 - 2 2 2 2 - = - = = = - + = + = α At t = 0.50 s the velocity is 29 o below the horizontal. v
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## This homework help was uploaded on 04/10/2008 for the course PHYSICS 125 taught by Professor Unknown during the Fall '08 term at University of Michigan.

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2-D and projectile motion Answers - Physics 125...

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