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

Answer:

It takes 2.04s to get to the maximum height

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:

• $v_{o} = 20\dfrac{m}{s}$

• $g = -9.8\dfrac{m}{s^{2}}$ (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:

$v_{f}=v_{o}+g*t$

$v_{f}-v_{o}=g*t$

$\dfrac{v_{f}-v_{o}}{g}=t$

$\dfrac{0-20}{-9.8}=t$

$t=2.04s$