Question

A bullet is fired into the air with an initial velocity of 1150 feet per second at an angle of

55o from the horizontal. The horizontal and vertical components of the velocity vector are:

Answer 1

Horizontal component: 660
`(ft)/(sec)`

Vertical component of the velocity = 942
`(ft)/(sec)`

Explanation:

Consider a right-angled triangle ABC (image is attached)where:

side AC = hypotenuse =
`1150(ft)/(ysec)`

side AB = adjacent side = horizontal component of velocity

side BC = opposite side = Vertical component of velocity

∠A = θ = 55°

Trigonometric functions will be applied:

1. sinθ =
`(opposite)/(hypotenuse)`

sin55° =
`(SideBC)/(1150)`

Cross-multiplication is appled:

Side BC =(1150
`(ft)/(sec)`)(sin55°)

BC = Vertical component of the velocity = 942
`(ft)/(sec)` (Rounded to 3 significant figures)

2. cosθ =
`(adjacent)/(hypotenuse)`

cos55° =
`(SideAB)/(1150)`

Cross-multiplication is appled:

Side AB =(1150
`(ft)/(sec)`)(cos55°)

AB = Horizontal component of the velocity = 660
`(ft)/(sec)` (Rounded to 3 significant figures)