Question

The velocity, V, of a specific object being dropped from a particular height, h, can be found by the radical function V = sqrt 2ha. Where A represents the acceleration in ft/sec2. If the acceleration due to gravity is 32.2 ft/sec2, at what height should you drop an object in order for it to have a velocity of 60 ft/sec?

0.9 feet
55.6 feet
55.9 feet
3,864 feet

Answer 1

YUP IT IS 55.9

Explanation:

I don't got an explanation, but I can say that I got 100 on my quiz.

Answer 2

55.9 feet

Explanation:

Basically you can just plug in what they give you

givens:

V=60 ft/sec

a= 32.2 ft/sec^2

function to find velocity:
`V=\sqrt{2ha`

if you plug everything in:
`60=\sqrt{2(h)(32.2)`

you can just put that equation into a website and ask it to solve for h which would give you your answer

but, if you want the work,

FIRST square both sides of the equation to get rid of the square root symbol on the left so,

`60^(2) =√(2(t)(32.2)) ^(2)`

which cancels out to

`3600= 2(t)(32.2)`

NEXT: divide both sides by 32.2 to get 2t by its self so,

`(3600)/(32.2) =(2(t)(32.2))/(32.2)`

which should get you to

`111.80124223=2(t)`

LAST: divide both sides by 2 to get t by itself so

`(111.80124223)/(2) =(2(t))/(2)`

and you should end up with

55.900621 = t

which rounded would be an answer of

t= 55.9 feet