1999MT1s

# 1999MT1s - Phys 125 Test 1 Fall 1999 Solutions 1 a(5 points...

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Unformatted text preview: Phys 125 Test 1 Fall 1999 Solutions 1 a) (5 points). 2 2 1 gt L = where t = 1 sec. ∴ L = 4.9 m b) (5 points) Now t 1 = 2 sec. = 2 2 1 gt 4L But we must subtract off the distance of the first ball to ground since we have calculated the distance to the ground of the second ball and we are asked for the distance between balls. L 1 = 4L – L = 3L L 1 = 14.7 m c). (10 points) Now t 2 = 3 sec and t 3 = 4 sec. Following the same method as above we find; L 2 = 9L – 4L = 5L L 2 = 24.5 m And L 3 = 16L – 9L = 7L L 3 = 34.3 m 2 a) (10 points) By looking at the figure on the test it is clear that tan θ = h/L The following equations of motion will be used in parts b) and c). For the bullet x B = v cos θ t y B = v sin θ t - 2 2 1 gt For the monkey x M = L y M = h - 2 2 1 gt b) (15 points) The bullet hits the monkey at x = L and y = 0 when the monkey has fallen the distance h. Therefore the time taken is given by g h t 2 = and thus we have at x =L g h V L 2 cos θ = h g L V MIN 2 cos θ = One can come up with other versions of this equation all related by tan...
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1999MT1s - Phys 125 Test 1 Fall 1999 Solutions 1 a(5 points...

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