T1a2s - Assignment 2 solutions (Fall term Physics 102) 2.57...

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2.57 The falling ball moves a distance of ( ) 15 m h before they meet, where h is the height above the ground where they meet. Apply 2 1 2 i yv t a t ∆= + , with to obtain a =− g () 2 1 15 m 0 2 hg −− = t , or 2 1 15 m 2 t . (1) Applying 2 1 2 i t a t + to the rising ball gives 2 1 25 m s 2 ht g t . (2) Combining equations (1) and (2) gives 15 m 25 m s t = = 0.60 s . Assignment 2 solutions (Fall term Physics 102) 2.62 (a) The velocity with which the first stone hits the water is 2 2 11 2 mm 2 2.00 2 9.80 50.0 m 31.4 ss fi vv a y ⎛⎞ + + ⎜⎟ ⎝⎠ m s . The time for this stone to hit the water is ( ) 1 2 31.4 m s 2.00 m s -9.80 m s t a ⎡⎤ ⎣⎦ == = 3.00 s . (b) Since they hit simultaneously, the second stone which is released 1. later, will hit the water after an flight time of 2.00 s. Thus, 00 s ( )( ) 2 2 2 2 2 2 50.0 m 9.80 m s 2.00 s 2 2 2.00 s i ya t v t ∆− = 15.2 m s . (c) From part (a), the final velocity of the first stone is 1 f v = 31.4 m s . The final velocity of the second stone is ( ) ( ) 2 222 15.2 m s 9.80 m s 2.00 s vva t =+= + = 34.8 m s .
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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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T1a2s - Assignment 2 solutions (Fall term Physics 102) 2.57...

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