# Evaluating the angle theta using inverse sin

In summary, to find the angle Θ in sin2Θ = 0.51, you can use the formula Θ = arcsin(0.51). To maximize the range of a projectile, it must leave the horizon at an angle of 45°. This means that sin 2θ = 1, and using the equation R = (Vi^2 * sin2Θ)/g, we can find the degree at which the projectile left the horizon.
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

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*}

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.

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 $$g\sin 2\theta = \frac{160}{56^2}\,\mathrm{m}\,\mathrm{s}^{-2} = \frac{5}{98}\,\mathrm{m}\,\mathrm{s}^{-2}.$$ 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 $$g\sin 2\theta = \frac{160}{56^2}\,\mathrm{m}\,\mathrm{s}^{-2} = \frac{5}{98}\,\mathrm{m}\,\mathrm{s}^{-2}.$$ 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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