My Math Forum [ASK] Shooting Target

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 March 13th, 2018, 02:27 AM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 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: 2,885 Thanks: 1504
March 30th, 2018, 12:40 PM   #3
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Quote:
 Originally Posted by Monox D. I-Fly 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.
Since you have calculated that $\displaystyle t= \frac{5\cos(\alpha)}{3}$. replace t by that in the second equation: $\displaystyle 10= (\frac{5\cos(\alpha)}{3})60 \sin(\alpha)- \frac{5\cos(\alpha)}{3})$.

Last edited by Country Boy; March 30th, 2018 at 12:45 PM.

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