Question

A baseball is thrown up in the air from a height of 3 feet with an initial velocity of 23 feet per second. What is the maximum height of the baseball and when does it reach this height?

[Use the equation h = -16t2 + v0t + h0.]

Answer 1

For this one we want to find the derivative of the position function you were given. The derivative is actually just the velocity function:

Derivative: v=-32t+v0

The initial velocity is 23ft/s, so v0=23. Now we need to figure out a method to finding that point where the ball is at its highest. If we think about it, the ball will be stationary when it reaches the max height, so v will be 0 then. We need to find t so that v=0:

0=-32t+23
32t=23
t=23/32

This gives us the amount of time it takes to reach max height, but now we need to find what that height actually is. The original position function will tell us that if we plug in the t we just solved for:

h=-16(23/32)^2+(-23)(23/32)+3
h=27.797ft

That's about it. If there's anything you have questions about feel free to ask! Also, you might want to double check that last part, the computer calculator is annoying to use >.>