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.

initial position d₀=0,

The maximum height at the motion with initial velocity v₀=20 m/s
is
H=v₀²/2g=20²/2•9.8 = 20.4 m (!!!)

H=v₀t-gt²/2,
v(fin)= v₀ -gt,
v(fin)=0, =>
0= v₀ -gt =>
v₀ =gt,
H=gt²-gt²/2= gt²/2,
t=sqrt{2H/g}=sqrt(2•20.4/9.8)=2.04 s

I don't get what d0 means. Explain properly

Displacement with Constant Acceleration.This equation may be used to find the displacement of any object moving with constant acceleration. that's what it says

To determine how long it takes for the object to reach the maximum height, we can use the kinematic equation for motion in the vertical direction:

vf = vo + at

Where:
vf = final velocity
vo = initial velocity
a = acceleration
t = time

In this case, the object is thrown directly upward, so the acceleration due to gravity acts in the negative direction. Therefore, the acceleration will be -9.8 m/s^2.

At the maximum height, the object's final velocity will be zero, since it momentarily comes to a stop before reversing direction.

So we can rewrite the equation as:

0 = 20.0 m/s - 9.8 m/s^2 * t

To solve for time (t), rearrange the equation:

9.8 m/s^2 * t = 20.0 m/s

t = 20.0 m/s / 9.8 m/s^2

t ≈ 2.04 seconds

Therefore, it takes approximately 2.04 seconds for the object to reach the maximum height of 24.0 m.