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Alenkinab [10]
3 years ago
7

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

Physics
1 answer:
babymother [125]3 years ago
4 0

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.

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