Question

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)

Answer 1

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:

$P(t)=gt^2 + vot + Po$

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 :

$P(t)=-32t^2 + 40t+ 5$

Firstly we find the time taken to reach its maximum height

Put, dP(t)/dt = 0

$\dfrac{d(-32t^2+40t+5)}{dt}=0\\\\-64t+40=0\\\\t=0.625\ s$

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

$P(0.625)=-32(0.625)^2 + 40(0.625)+ 5\\\\=17.5\ \text{feet}$

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