ch02-p058 - 58. To find the "launch" velocity of...

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58. To find the “launch” velocity of the rock, we apply Eq. 2-11 to the maximum height (where the speed is momentarily zero) () 2 00 0 9.8 m/s 2.5 s vv g t v =− ¡ so that v 0 = 24.5 m/s (with + y up). Now we use Eq. 2-15 to find the height of the tower
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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