March 13th, 2018, 02:27 AM  #1 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry  [ASK] Shooting Target
A target is located at the distance 100 m and height 10 m. How much is the elevation angle of the shooter to be able to hit the target if the initial velocity is 60 m/s and g = $\displaystyle 10m/s^2$? What I have done: 100 = 60sin$\displaystyle \alpha$t and 10 = 60sin$\displaystyle \alpha$t  $\displaystyle \frac{1}{2}(10)t^2$ $\displaystyle t=\frac{5\cos\alpha}{3}$ and $\displaystyle 10=t(60\sin\alpha)t)$ Dunno what to do from here on. Last edited by skipjack; March 13th, 2018 at 04:07 PM. 
March 13th, 2018, 03:12 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,767 Thanks: 1422  
March 30th, 2018, 12:40 PM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  Quote:
Last edited by Country Boy; March 30th, 2018 at 12:45 PM.  

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