Question

A ball is thrown vertically into the air from the ground with an initial velocity of 96 ft/sec. Its height, in feet, as a function of time, in seconds, is given by: h(t)=96t-16t 2 .
Estimate the ball’s instantaneous velocity at time t = 3, including units. Explain what this value means in terms of the ball’s path.
Instantaneous velocity at time t = 3 is:
Intervals used and corresponding velocities:
???

Answer 1

let's use v for velocity, h is for the distance in the vertical plane.
we know \[v=\frac{ h }{ t }\]
correct?

Answer 2

OK...yeah

Answer 3

Now, how do we find the h at t=3?

Answer 4

I've been evaluating intervals into function at like 3.0 and 2.9?

Answer 5

Is there a different way?

Answer 6

what did you get for value of your distance?

Answer 7

Like 1.6 ft/sec @ [2.9,3] & .016 ft/sec @ [2.999,3]

Answer 8

From left to right it appears to be nearing zero?

Answer 9

h(3) = 96(3)- 16 (3*3) = 144 ft/sec.

Answer 10

Yeah

Answer 11

Oh....is that it?

Answer 12

I was taking the two intervals and dividing by the differnece. Getting zero

Answer 13

Like....F9x1_-f(x0)/x1-x0

Answer 14

Like....Fx1_-f(x0)/x1-x0

Answer 15

use \[v = \frac{ h }{ t }\]
Where h = 144 ft and t = 3 sec.

Answer 16

is the question from physics or introductory calculus?

Answer 17

Intro calc