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

#### Monoxdifly

##### Well-known member
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
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.

#### Monoxdifly

##### Well-known member
But why do things "get complicated"?
Probably just because I'm not used to the approximation of irrational numbers.