Question

Arocket launches at an angle of 33.6 degrees from the horizontal at a

Answer 1

Answer:

Y component = 32.37

Explanation:

Given:

Angle of projection of the rocket is, $\theta=33.6$

Initial velocity of the rocket is, $u=58.5$

A vector at an angle $\theta$ with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.

If a vector 'A' makes angle $\theta$ with the horizontal, then the horizontal and vertical components are given as:

$A_x=A\cos \theta(\textrm{Horizontal or X component})\\A_y=A\sin \theta(\textrm{Vertical or Y component})$

Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:

$u_y=u\sin \theta$

Plug in the given values and solve for $u_y$. This gives,

$u_y=(58.5)(\sin 33.6)\\u_y=58.5\times 0.55339\\u_y=32.373\approx32.37(\textrm{Rounded to two decimal places})$

Therefore, the Y component of initial velocity is 32.37.