Patrick F. answered • 10/15/15

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a) The formulas are

v

_{y0}= v_{0}sin θv

_{x0}= v_{0}cos θPlugging in v

_{0}= 40 m/s and θ = 37 deg, we havev

_{y0}= 24.07 m/sv

_{x0}= 31.95 m/sb) When the arrow reaches its highest point, its vertical velocity (v

_{y}) is zero. The formula isv

_{y}^{2}= v_{y0}^{2}+ 2a_{y}ΔyThe point where v

_{y}= 0 corresponds to a specific Δy (the height the arrow goes relative to the archer on top of the castle); v_{y0}was found in part (a), and a_{y}= -g.Δy = v

_{y0}^{2}/ 2g = (24.07 m/s)^{2}/ (2*9.8 m/s^{2}) = 29.56 mMeasured relative to the ground, the arrow reaches a height of 10 m + 29.56 m = 39.56 m.

c) Since the final y position of the arrow is different from its initial position, the process for finding this is a bit more tedious, but still possible. I'll just quote the answer: https://en.wikipedia.org/wiki/Range_of_a_projectile. Using my notation, this is

Δx = (v

_{x0}/ g)(v_{y0}+ sqrt[v_{y0}^{2}+ 2gy_{0}]where y

_{0}= 10 m. Plugging in the values gives 169.26 m.