Question

Izzy hits a ball of a tee 2 feet off the ground, with an initial velocity of 64 feet per second. What will be the maximum height the ball reaches?

Answer 1

Answer:

Step-by-step explanation:

I figured this out with calculus since it's easier that way. The position function for the ball is

$s(t)=-16t^2+64t+2$. The first derivative of position is velocity, so we need to find the first derivative of the position function which is

v(t) = -32t + 64

Now, where the ball is at its highest point is where the velocity is equal to 0, so setting the velocity function equal to 0 allows us to determine how many seconds it takes to get to that max height.

0 = -32t + 64 and

-64 = -32t so

t = 2 seconds. It takes 2 seconds to get to its max height. In order to determine that max height, we sub 2 in for t in the position function:

$s(2)=-16(2)^2+64(2)+2$ and

s(2) = 66 feet. The max height of the ball is 66 feet.