p34_063 - ◦ = − 8 7 m s which means that the distance...

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63. (a) The Sun is far enough away that we approximate its rays as “parallel” in this Figure. That is, if the sunray makes angle θ from horizontal when the bird is in one position, then it makes the same angle θ when the bird is any other position. Therefore, its shadow on the ground moves as the bird moves: at 15 m/s. (b) If the bird is in a position, a distance x> 0 from the wall, such that its shadow is on the wall at a distance 0 y h from the top of the wall, then it is clear from the Figure that tanθ = y/x . Thus, dy dt = dx dt tan θ =( 15 m / s) tan 30
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Unformatted text preview: ◦ = − 8 . 7 m / s , which means that the distance y (which was measured as a positive number downward from the top of the wall) is shrinking at the rate of 8 . 7 m/s. (c) Since tan θ grows as 0 ≤ θ < 90 ◦ increases, then a larger value of | dy/dt | implies a larger value of θ . The Sun is higher in the sky when the hawk glides by. (d) With | dy/dt | = 45 m/s, we ±nd v hawk = ¯ ¯ ¯ ¯ dx dt ¯ ¯ ¯ ¯ = ¯ ¯ ¯ dy dt ¯ ¯ ¯ tan θ so that we obtain θ = 72 ◦ if we assume v hawk = 15 m/s....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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