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

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To determine the velocity needed for a rock shot from a slingshot at a 37-degree angle to hit a bird at coordinates (50, 8) m, the equations of motion were applied. After substituting known variables, the time of flight was calculated to be approximately 2.43 seconds. This led to the equation for horizontal distance, resulting in a required initial velocity of about 25.8 m/s. The discussion highlights the challenge of working with irrational numbers in calculations. The complexity arises from the need to accurately approximate these values in the context of projectile motion.
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.