Question

An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. Determine how long it takes to get to the maximum height of 24.0 m.

Answer 1

t=d/v
t=24m/20m/s
t=6/5 s
t=1.2s

Answer 2

It takes 2.04s to get to the maximum height

Step-by-step explanation:

This is a vertical throw problem so it can be treated as a uniform accelerated rectilinear motion. For computing time we are going to use the formula:

`v_(f)=v_(o)+g*t`

where
`v_(f)` is the final velocity,
`v_(o)` is the initial velocity,
`t` is the time and,
`g` is the gravity.

For solving this kind of problems we need at least three values. The values we have are:

• 
  `v_(o) = 20(m)/(s)`

• 
  `g = -9.8(m)/(s^(2))` (negative because gravity's direction is oposite from the object's moving direction)
• 
  `v_(f)=0` (final velocity equals zero because at maximun height the object stops moving)

Now:

`v_(f)=v_(o)+g*t`

`v_(f)-v_(o)=g*t`

`(v_(f)-v_(o))/(g)=t`

`(0-20)/(-9.8)=t`

`t=2.04s`