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Jinnie
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
abg velocity would be (h(2) - h(2.01))/(2-2.01)
how do i find the instantaneous velocity at two seconds and the direction of the water balloon
it involves calculus. do you know what a derivative is?
instantaneous vel would be dh/dt (differential wrt t) ...
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
alright let me enter this into my calc and see
stay with me please. i need to know this question
am i plugging 2 into t^2 and 2.01 into 30t
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?
with calc edgeoftheearth meant means calculus .. not calculator.. :)
do you have exposure in calculus?
am i familiar with calculus?
when i plug in 2, i get 196.3 and 2.01, I get 195.6584
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
ok so i subtract 195.6584 from 196.3 and then divide by .01?
Yes, but your change in height is a little off.. h(2.01) = 195.6584 and h(2) = 196. 195.6584 - 196 = -0.3416.
so for the instantaneous velocity, what must i do? i didnt understand you last time
for instantaneous velocity you have to take the derivative of the position function (h(t)).