Question

An object is thrown vertically and has a speed of 25 m/s when it reaches 1/4 of its maximum height above the ground (assume it starts from ground level). What is the original launch speed of the object?

Answer 1

`v_(i) =28.86(ft)/(s)`

Step-by-step explanation:

Conceptual analysis

We apply the kinematic formula for an object that moves vertically upwards:

`(v_(f) )^(2) =(v_(i) )^(2) -2*g*y`

Where:

`v_(f)` : final speed in ft/s

`v_(i)` : initial speed in ft/s

g: acceleration due to gravity in ft/s²

y: vertical position at any time in ft

Known data

For
`v_(f) = 25(ft)/(s)` ,
`y=(1)/(4) h`; where h is the maximum height

for y=h,
`v_(f) =0`

Problem development

We replace
`v_(f) = 25(ft)/(s)` ,
`y=(1)/(4) h (ft)` in the formula (1),

[
`25^(2) =(v_(i) )^(2) -2*g*(h)/(4)` Equation (1)

in maximum height(h):
`v_(f) =0`, Then we replace in formula (1):

`0=(v_(i) )^(2) - 2*g*h`

`2*g*h=(v_(i) )^(2)`

`h=((v_(i))^(2)  )/(2g)` Equation(2)

We replace (h) of Equation(2) in the Equation (1) :

`25^(2) =(v_(i) )^(2) -2g(((v_(i))^(2)  )/(2g) )/(4)`

`25^(2) =(v_(i) )^(2) -((v_(i))^(2)  )/(4)`

`25^(2) =(3)/(4) (v_(i) )^(2)`

`v_(i) =\sqrt{(25^(2)*4 )/(3) }`

`v_(i) =28.86(ft)/(s)`