MHB [ASK] How much velocity is needed for the rock to be able to hit that bird?

Monoxdifly
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A bird is located at the (50, 8 ) m coordinate. A boy shot a rock at it using a slingshot with the elevation angle $$37^{\circ}$$. How much velocity is needed for the rock to be able to hit that bird?

I substituted all known variables (including the gravity acceleration $$10m/s^2$$, with $$\sin37^{\circ}=\frac{3}{5}$$ and $$\cos37^{\circ}=\frac{4}{5}$$) to $$x=v_0\cos\alpha t$$ and $$y=v_0\sin\alpha t-\frac{1}{2}gt^2$$. Substituting the $$v_0$$ I got from both equations resulted in $$\frac{125}{2t}=\frac{40+25t^2}{3t}$$ and I got $$t^2=5,9$$. Am I right? Because if it is indeed the right value of t, everything will be complicated from there.
 
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Yes, that is correct and then t= \sqrt{5.9}= 2.43 (to two decimal places). But why do things "get complicated"? You have x= 50= v_0 cos(37)(2.43)= 1.94 v_0. v_0= 50/1.94= 25.8 m/s.
 
Country Boy said:
But why do things "get complicated"?

Probably just because I'm not used to the approximation of irrational numbers.
 
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