for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For what other launch angle will the projectile have the same range?

Respuesta :

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

[tex]R=\dfrac{u^2\ \sin2\theta}{g}[/tex]

Here,

u is the speed of launch of projectile

Here, [tex]\theta=30^{\circ}[/tex]

We need to find the other launch angle when the projectile have the same range, such that,

[tex]\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}[/tex]

[tex]\sin(60)=\sin2\alpha[/tex]

[tex]\dfrac{\sqrt3}{2}=\sin2\alpha[/tex]

[tex]\alpha =30^{\circ}[/tex]

So, the other angle is 30 degrees. Hence, this is the required solution.