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

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

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