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Eclair_de_XII
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Homework Statement
"Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water (Fig. 4-38). Although the fish sees the insect along a straight-line path at angle Φ and distance d, a drop must be launched at a different angle θ0 if its parabolic path is to intersect the insect. If Φ = 36.0° and d = 0.900 m, what launch angle θ0 is required for the drop to be at the top of the parabolic path when it reaches the insect?"
Homework Equations
##y=(tanθ_0)x-\frac{gx^2}{2(v_0cosθ_0)^2}##
##R = \frac{v_0^2}{g}sin2θ_0##
##θ=36°##
##|d|=0.9m##
The Attempt at a Solution
##\vec d = (0.728m)i+(0.529m)j##
##0.728m=\frac{v_0^2}{9.8\frac{m}{s^2}}sin72°##
##7.1344=v_0^2(sin72°)##
##v_0=2.335\frac{m}{s}##
##0.529m=(0.72812)(tanθ)-\frac{(9.8\frac{m}{s^2})(0.728m)^2}{2[(2.335\frac{m}{s})(cosθ°)]^2}sin72°##
##(-0.729)(sec^2θ)+(0.72812)tanθ-0.529=0##
##(-0.729)(tan^2θ+1)+(0.72812)tanθ-0.529=0##
##(-0.729)tan^2θ+(0.72812)tanθ-1.257=0##
Let:
##q=tanθ##
so:
##q^2-0.999q+1.724=0##
And then I come to an impasse at the 4ac>b2 block. I'm thinking I'm mistaking my R for my x... So how would I find R for v0?
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