March 12th, 2018, 09:23 PM  #1 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,825 Thanks: 125 Math Focus: Trigonometry and Logarithm  [ASK] Shot Bird
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. 
March 13th, 2018, 02:23 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,724 Thanks: 1378 
If the shot originates at the origin, then your solution for time is correct. You may want to try this ... $\Delta x = 50$ and $\Delta y = 8$ the idea here is to remove the parameter of time ... $\Delta x = v_0 \cos{\theta} \cdot t \implies t = \dfrac{\Delta x}{v_0 \cos{\theta}}$ substitute the above value for time in the equation $\Delta y = v_0 \sin{\theta} \cdot t  \dfrac{1}{2}gt^2$ $8 = v_0 \sin(37) \cdot \dfrac{50}{v_0 \cos(37)}  5 \left[\dfrac{50}{v_0 \cos(37)}\right]^2$ $\sin(37) \approx \dfrac{3}{5}$ and $\cos(37) \approx \dfrac{4}{5}$ ... $8 = \dfrac{75}{2}  \dfrac{78125}{4v_0^2}$ your solution should be $v_0 \approx 25.7 \, m/s$ 
March 13th, 2018, 04:03 PM  #3 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,825 Thanks: 125 Math Focus: Trigonometry and Logarithm 
I see. Damn physics, it's kinda hard to get an integer answer...


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