Question

A girl standing on the top of a building 200 feet highs throws a water balloon up in the air at a velocity of 30ft/sec. The equation for the height of the water balloon is h(t)=-16t^2 +30t + 200

Find the average velocity of the water balloon between 2 sec and 2.01 sec

Answer 1

abg velocity would be (h(2) - h(2.01))/(2-2.01)

Answer 2

how do i find the instantaneous velocity at two seconds and the direction of the water balloon

Answer 3

it involves calculus. do you know what a derivative is?

Answer 4

instantaneous vel would be dh/dt (differential wrt t) ...

Answer 5

direction of baloon would be the sign of the instantaneous velocity ..if it is positive then it would be in positive direction of the axis you are measuring your height anf vice versa

Answer 6

alright let me enter this into my calc and see

Answer 7

stay with me please. i need to know this question

Answer 8

am i plugging 2 into t^2 and 2.01 into 30t

Answer 9

no mate you plugging both into the h(t) function ... its like you finding the value of h(t) at 2 and again value of h(t) at 2.01

what class are you in?

Answer 10

calc

Answer 11

ok i see

Answer 12

with calc edgeoftheearth meant means calculus .. not calculator.. :)

Answer 13

i dont get you

Answer 14

do you have exposure in calculus?

Answer 15

am i familiar with calculus?

Answer 16

when i plug in 2, i get 196.3 and 2.01, I get 195.6584

Answer 17

does that make sense? you are basically finding the change in height (ft) from 2 to 2.01 seconds, and then dividing that by the change in time, .01 s

Answer 18

ok so i subtract 195.6584 from 196.3 and then divide by .01?

Answer 19

Yes, but your change in height is a little off.. h(2.01) = 195.6584 and h(2) = 196. 195.6584 - 196 = -0.3416.

Answer 20

i see

Answer 21

so for the instantaneous velocity, what must i do? i didnt understand you last time

Answer 22

for instantaneous velocity you have to take the derivative of the position function (h(t)).

Answer 23

i got zero, am i right?