TestMidterm1-sec1-v3

# TestMidterm1-sec1-v3 - Midterm 1 FAJANS Solution of Midterm...

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Solution of Midterm 1 (by Haoyu) Solution 1 Set the original position of the lion to be x = 0. The position for the lion x lion would be: x lion = v 0 t (1) and the position for the gazelle x gazelle would be: x gazelle = d + 1 2 at 2 (2) Therefore for the lion to catch the gazelle we have equation: x lion = x gazelle v 0 t = d + 1 2 at 2 1 2 at 2 - v 0 t + d = 0 (3) If the gazelle want to escape, then it must make sure that the above equation has no position solutions. That is: v 2 0 - 4( 1 2 a ) d < 0 (4) Solve this inequality we get a > v 2 0 2 d (5) Therefore, the maximum acceleration that the gazelle would still be caught is v 2 0 2 d Solution 2 Let the x -distance between the monkey and the hunter be d . Since the riﬂe is aimed at the monkey we have: d tan θ = h (6) Set the origin at the hunter and let the initial speed of the bullet be v 0 , the projectile trace of the bullet is: x bullet = v 0 t cos θ (7) y bullet = v 0 t sin θ - 1 2 gt 2 (8) The free fall trace of the monkey is:

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## This note was uploaded on 10/25/2010 for the course PHYSICS 4fafas taught by Professor Golightly during the Spring '10 term at Berkeley.

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TestMidterm1-sec1-v3 - Midterm 1 FAJANS Solution of Midterm...

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