# T1a3s - Assignment 3 solutions(Fall term Physics 102 3.29...

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3.29 We choose our origin at the initial position of the projectile. After 3.00 s, it is at ground level, so the vertical displacement is yH ∆ =− . To find H , we use 2 1 2 iy y yv t a t ∆= + , which becomes () ( ) ( ) ( 2 2 1 15 m s sin 25 3.0 s 9.80 m s 3.0 s 2 H −= ° + − ) , or H = 25 m . Assignment 3 solutions (Fall term Physics 102) 3.55 (a) The time to reach the fence is 130 m 159 m cos 35 ix i i x t vv v == = ° . At this time, the ball must be 20 m above its launch position. 2 1 2 iy y t a t + gives ( ) 2 2 159 m 159 m 20 m sin 35 4.90 m s i ii v ⎛⎞ =° − ⎜⎟ ⎝⎠ . From which, i v = 42 m s . (b) From above, 159 m 159 m 42 m s i t v == = 3.8 s (c) ( ) 42 m s cos 35 xi x ° = 34 m s ( ) ( ) ( ) 2 42 m s sin 35 9.80 m s 3.8 s yi yy vva t =+= ° = 13 m s 22 3 4 .1 ms 1 3 .4 xy vvv =+ = + = 37 m s 4.17 From , , or 0 x F Σ= 12 cos 30.0 cos60.0 0 TT °− °= ( ) 21 1.73 T = T . (1) Then becomes 0 y F ( ) 11 sin 30.0 1.73 sin 60.0 150 N 0 °+ = , which gives 1 T = 75.0 N in the right side cable . Finally, Equation (1) above gives 2 T = 130 N in the left side cable .

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4.27 (a) The resultant external force acting on this system having a total mass of 6.0 kg is 42 N directed horizontally toward the right. Thus, the acceleration
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## This note was uploaded on 01/15/2011 for the course PHYS 102 taught by Professor Vandenberg during the Fall '08 term at University of Victoria.

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T1a3s - Assignment 3 solutions(Fall term Physics 102 3.29...

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