Phy121f10 Midterm 1 Solutions

# Phy121f10 Midterm 1 Solutions - Phy 121 Fall 2010 Midterm 1...

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Phy 121 Fall 2010 Midterm 1 Solutions: Form 0 1. 2. 21 10 1 1.5 / 93 xx v m s tt  22 00 2 2 7 1 0.714 / ; 9 2 2 1 3 7 (6 2) ( 0.714) (6 2) 25.3 2 a m s x x v t a t m     3. 3 for very small : |3t|>|5 | initial motion is backward; maximum power of is 3 is not constant; ta 4. There is vacuum on the surface of the moon and all objects fall with the same speed. Both equations for t and v below do not contain any mass. Thus the mass doesn’t matter . Since / 1/ 6 moon earth gg and 2 1 2 2 2 = 6 6 2 / 6 moon earth h h h h g t t t t g g g   . With 11 6 6 6 monn earth earth earth v g t v g t v . 5. The velocity with which ball 1 arrives at the launch height on its way down is equal to the launch velocity, but pointing downward. Ball 2 must be thrown downward with the same velocity. The launch velocity of ball 1 is: 0 2 ( ) 2 ( ) 2 9.81 (4.0 1.5) 7.00 / top launch top launch v v g h h v g h h ms 6.

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## This note was uploaded on 03/06/2011 for the course PHY 121 taught by Professor Stephens during the Fall '08 term at SUNY Stony Brook.

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Phy121f10 Midterm 1 Solutions - Phy 121 Fall 2010 Midterm 1...

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