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

Monoxdifly

Well-known member
Aug 6, 2015
284
A bird is located at the (50, 8 ) m coordinate. A boy shot a rock at it using a slingshot with the elevation angle \(\displaystyle 37^{\circ}\). How much velocity is needed for the rock to be able to hit that bird?

I substituted all known variables (including the gravity acceleration \(\displaystyle 10m/s^2\), with \(\displaystyle \sin37^{\circ}=\frac{3}{5}\) and \(\displaystyle \cos37^{\circ}=\frac{4}{5}\)) to \(\displaystyle x=v_0\cos\alpha t\) and \(\displaystyle y=v_0\sin\alpha t-\frac{1}{2}gt^2\). Substituting the \(\displaystyle v_0\) I got from both equations resulted in \(\displaystyle \frac{125}{2t}=\frac{40+25t^2}{3t}\) and I got \(\displaystyle t^2=5,9\). Am I right? Because if it is indeed the right value of t, everything will be complicated from there.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
469
Yes, that is correct and then [tex]t= \sqrt{5.9}= 2.43[/tex] (to two decimal places). But why do things "get complicated"? You have [tex]x= 50= v_0 cos(37)(2.43)= 1.94 v_0[/tex]. [tex]v_0= 50/1.94= 25.8[/tex] m/s.
 

Monoxdifly

Well-known member
Aug 6, 2015
284