Question

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

velocity of 58.5, calculate the Y component of the initial velocity vector of
the rocket.
Round your answer to two decimal places (XX.XX), NO UNITS!!
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Answer 1

Y component = 32.37

Step-by-step 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* 0.55339\\u_y=32.373\approx32.37(\textrm{Rounded to two decimal places})`

Therefore, the Y component of initial velocity is 32.37.