A baseball is thrown in a parabolic arc. It's position above the ground at a given point in time can be represented by the quadratic function
p(t) = {gt2 + vot + Po, where t > 0. g is -32 ft/sec/sec, vo is initial
velocity, and Po is its initial position above the ground. If the ball was thrown
straight up at 40 ft/sec when it was 5 ft above the ground, how high did it go?
(Type the number, do not include units in your answer)

P(t) = 17.5 feet

Step-by-step explanation:

The position above the ground at a given  point in time can be represented by the quadratic function as follows:

Where,

Po is the initial position above the ground

If the ball was thrown  straight up at 40 ft/sec when it was 5 ft above the ground.

We need to find the height reached by the ball.

The equation will become :

Firstly we find the time taken to reach its maximum height

Put, dP(t)/dt = 0

Now put t = 0.625 in equation (1) as follows :

So, it will go at a height of 17.5 feet.

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step-by-step explanation: