Question

A bullet is shot straight up in the air with an initial velocity of v=1000 feet per second. If the equation describing the bullets height from the ground is given by f(t)=−16t2+1000t+7, over what interval is the bullet speeding up, and over what interval is it slowing down?

Answer 1

It slows down from 0-31.25s

It speeds up from 31.25-62.507s

Step-by-step explanation:

If we find the maximum of the equation, we will know the moment when it changes direction. It will slow down on the first interval, and speed up on the second one.

`Y = -16t^2+1000t+7`

`Y' = -32t+1000 = 0` Solving for t:

t = 31.25s

The bullet will slow down in the interval 0-31.25s

Let's now find the moment when it hits ground:

`Y = 0 = -16t^2+1000t+7` Solving for t:

t1 = -0.0069s t2 = 62.507s

The bullet will speed up in the interval 31.25-62.507s