Johnny Blade
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What is the angle if the reach is three (3) times the maximum height.
I put both of the reach and max height formula together to isolate theta
p=\frac{V_{0}^{2}sin(2\theta)}{-g} and h_{max}=\frac{V_{0}^{2}sin^{2}\theta}{-2g}
At the end it gave me this:
\frac{3sin\theta}{2}=2cos\theta
Am I getting rusty on my trigonometry? How would you solve this?
I put both of the reach and max height formula together to isolate theta
p=\frac{V_{0}^{2}sin(2\theta)}{-g} and h_{max}=\frac{V_{0}^{2}sin^{2}\theta}{-2g}
At the end it gave me this:
\frac{3sin\theta}{2}=2cos\theta
Am I getting rusty on my trigonometry? How would you solve this?