The function h(t) = −16t2 + 32t + 24 models the height h, in feet, of a ball t seconds after it is thrown straight up into the air. What is the initial velocity and the initial height of the ball?

Answer 1

To find the initial height, we want to find the value of h when t=0. To do this substitute 0 for t and solve for h.

h(t)= -16t² + 32t + 24

h(0) = -16(0)² + 32(0) + 24

h(0) = 24

The initial height is 24 feet.

In general the height is represented by the constant in the equation. In this case, 24.

The initial velocity is represented by the coefficient of x, in this case 32.

The initial velocity is 32 feet/second. 

Answer 2

I'm guessing this problem is for calculus and not for algebra.

So, taking the derivative of displacement yields velocity and d/dx of velocity yields acceleration.

h(t) = -16t² +32t +24, thus, h´(t) = 32t + 32, when t = 0, right at the moment the ball leaves that persons hand, you get v₀ (initial velocity). Plug in 0 (into the derivative, h´(t)) and label with ft/s.

The initial height is really easy: just plug 0 into the original equation (h(t)). That one is super easy and make to label with feet.