Evaluating the angle theta using inverse sin

[Mohmmad Maaitah]

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

 

[DrClaude]

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##:
$$
\begin{align*}
\sin 2\theta &= 0.51 \\
\arcsin( \sin 2\theta ) &= \arcsin(0.51) \\
2\theta &= \arcsin(0.51) \\
\theta &= \frac{\arcsin(0.51)}{2}
\end{align*}
$$

 

[Lnewqban]

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.

 

[pasmith]

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 [tex]
g\sin 2\theta = \frac{160}{56^2}\,\mathrm{m}\,\mathrm{s}^{-2} = \frac{5}{98}\,\mathrm{m}\,\mathrm{s}^{-2}.[/tex] Using [itex]g = 9.8\,\mathrm{m}\,\mathrm{s}^{-2}[/itex] then gives [itex]\sin 2\theta = \frac12[/itex], and [itex]\sin 30^{\circ} = \frac12[/itex] is one of the results which you should know.

 

[SammyS]

Note that [tex]
g\sin 2\theta = \frac{160}{56^2}\,\mathrm{m}\,\mathrm{s}^{-2} = \frac{5}{98}\,\mathrm{m}\,\mathrm{s}^{-2}.[/tex] Using [itex]g = 9.8\,\mathrm{m}\,\mathrm{s}^{-2}[/itex] then gives [itex]\sin 2\theta = \frac12[/itex], and [itex]\sin 30^{\circ} = \frac12[/itex] 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##