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Unformatted text preview: can be o
solved to give ymax = v2
o
. We have two cases with different initial velocities. 2g v 21
v 22
o
For case 1, ymax 1 = for case 2, ymax 2 = o . Dividing the first equation by 2g
2g the second gives ymax 1
ymax 2 v 21
o
2
v 21
2g v o1
o
= 2 = 2 . So ymax 1 = 2 ymax 2 v o2 v o2
v o2
2g PHY 2048 Spring 2014 – Acosta, Rinzler Exam 1 solutions 2 ⎛v ⎞
Hence the height of the first will be Q = 2 = ⎜ o1 ⎟ times the height of the 2nd. v o2 ⎝ v o2 ⎠
Since the speed with which the 1st was thrown was 100 m/s and the speed of the v 21
o 2 ⎛ 100 ⎞
2nd was 10 m/s , Q = ⎜
= 100. ⎝ 10 ⎟
⎠ Problem 9 A ball is thrown horizontally from the top of a cliff from a height h=23.0 m (or 32 m, 41 m) above the level plain below. It strikes the plain at an angle of 60° with respect to the horizontal. With what...
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This document was uploaded on 03/24/2014.
 Spring '14
 Physics

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