An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. how high does it rise (v = 0 cm/s at top of rise). remember, acceleration is -9.80 m/s2.

The maximum height is 20.4 meters

Explanation: We have that the initial velocity is 20m/s and the initial position is 0m.

The acceleration of the object is:

a(t) = -9.8m/s^2

for the velocity, we integrte over time.

v(t) = -9.8m/s^2*t + 20m/s

and the position is:

p(t) = -4.9m/s^2*t^2 + 20m/s*t + 0

And we want to do the maximum height (in this case the maximum value of p(t))

First, we need to find the time in wich te velocity is equal to zero, which means that the object stops to going up and starts moving down.

v(t) = 0 = -9.8m/s^2*t + 20m/s

t = (20/9.8)s = 2.04s

Now, we put this time in the position equation:

p(2.04s) = -4.9m/s^2*(2.04s)^2 + 20m/s*2.04s = 20.4m

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im guessing the west of station b