Evaluating the angle theta using inverse sin

AI Thread Summary
To find the angle Θ in the equation sin2Θ = 0.51, the correct approach is to use the inverse sine function. By setting 2Θ = arcsin(0.51), it follows that Θ = arcsin(0.51)/2. Additionally, the discussion touches on projectile motion, noting that to maximize range, the launch angle should ideally be 45°, where sin2Θ equals 1. A calculation involving gravitational acceleration (g) indicates that sin2Θ can also equal 0.5, corresponding to an angle of 30°. The thread emphasizes the importance of correctly applying trigonometric identities in these calculations.
Mohmmad Maaitah
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Homework Statement
Find the degree which the projectile left the horizon with.
Relevant Equations
I used R= (Vi^2 * sin2Θ)/g
I just need to know how to find Θ in sin2Θ=0.51
I know I can use Θ = arcsin(0.51)
but what about sin2Θ = 0.51
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The arcsin is just the inverse operation of sin, ##\arcsin( \sin (x)) = x##, so simply keep the factor 2, i.e., set ##x = 2\theta##:
$$
\begin{align*}
\sin 2\theta &= 0.51 \\
\arcsin( \sin 2\theta ) &= \arcsin(0.51) \\
2\theta &= \arcsin(0.51) \\
\theta &= \frac{\arcsin(0.51)}{2}
\end{align*}
$$
 
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Consider that to maximize the range (R), the projectile must leave the horizon with an angle of 45°.
At that angle, sin 2θ = sin [(2)(45°)] = 1.
 
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Mohmmad Maaitah said:
Homework Statement: Find the degree which the projectile left the horizon with.
Relevant Equations: I used R= (Vi^2 * sin2Θ)/g

I just need to know how to find Θ in sin2Θ=0.51
I know I can use Θ = arcsin(0.51)
but what about sin2Θ = 0.51
View attachment 327270

Note that <br /> g\sin 2\theta = \frac{160}{56^2}\,\mathrm{m}\,\mathrm{s}^{-2} = \frac{5}{98}\,\mathrm{m}\,\mathrm{s}^{-2}. Using g = 9.8\,\mathrm{m}\,\mathrm{s}^{-2} then gives \sin 2\theta = \frac12, and \sin 30^{\circ} = \frac12 is one of the results which you should know.
 
pasmith said:
Note that <br /> g\sin 2\theta = \frac{160}{56^2}\,\mathrm{m}\,\mathrm{s}^{-2} = \frac{5}{98}\,\mathrm{m}\,\mathrm{s}^{-2}. Using g = 9.8\,\mathrm{m}\,\mathrm{s}^{-2} then gives \sin 2\theta = \frac12, and \sin 30^{\circ} = \frac12 is one of the results which you should know.
Looks like a typo there.

Should be ##\quad \dfrac 1 g \sin 2\theta = \, \dots##

or ##\quad \sin 2\theta = g\dfrac{160}{56^2} \, \dots##
 
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