Question

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, it height in feet after t second is given by y = 70 t - 16 t^2. Find the average velocity for the time period beginning when t = 2 and lasting
(i) 0.1 seconds

Answer 1

{y(2.1) -y(2) } / {2.1 -2}

Answer 2

does that help?

Answer 3

-0.85?

Answer 4

hmm

Answer 5

find y(2.1) = 70*2.1 -16*(2.1)^2

Answer 6

find y(2) = 70*2 -16*(2)^2

Answer 7

find the difference

Answer 8

divide it by 2.1-2 = .1

Answer 9

let me know what you get.

Answer 10

4.4 :) I got it correct.! And if I wanted to find 0.01 seconds and 0.001 seconds. How would I solve that?

Answer 11

same basic set up
just use t=2 and t = 2.01 for your times
and then t=2 and t=2.001

Answer 12

{y(2.01) -y(2) } / {2.01 -2} etc.

Answer 13

Okay, I got those correct. Now, finally based on the results, how do I guess what the instantaneous velocity of the ball is when t =2?

Answer 14

what are your values nearing as the time interval gets smaller?
4.4 -> ? -> ? ->

Answer 15

4.4 ->5.84 ->

Answer 16

4.4 ->5.84 ->5.984 -> ?

Answer 17

6?

Answer 18

thank you!

Answer 19

yes:) yw!