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if a ball is thrown into the air with a velocity of 40ft/s, its height in feet t seconds later is given by y=40t-16t^2. find the average velocity for the time period beginning when t=2 and lasting 0.5 seconds

Mathematics
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y = the upward distance = 40t - 16t^2 distance/time = velocity Let V be the velocity so V = y /t = 40 - 16 t, t = 2 V = 40 - 32 = 8
ok wait
this is the equation y=40t-16t^2

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Other answers:

how did u plug t=2 and (0.5)seconds
yes, y = 40t - 16t^2 y /t is velocity = 40 - 16t st t = 2, v = 8
ok u factor
Vavg = (yf-yi)/(tf-ti) yf = 40(2.5)-16(2.5)^2 = 0 yi = 40(2)-16(2)^2 = 16 tf = 2.5 ti = 2 Vavg = (0-16)/(2.5-2) = -16/.5 = -32 ft/s
what is tha 2.5
The problem asked for AVERAGE velocity from t=2 and lasting .5 seconds, so the initial time is t=2 and the final time is t=2.5
ok how did u get 2.5 u added then
yes, if you start at t=2 and end .5 seconds later, then the final time is t=2.5
cool
p.kanan found the velocity when t=2, which is not the average velocity from t=2 to t=2.5 as the problem asked.
ohh ok
thank u quantish
do you understand? there is a difference between average velocity and velocity at a certain instant.
sip another qs. in the denominator is no supose to be 40(2)-16(0.5)^2
I really don't understand what you're asking. what denominator? your change in time i.e. time final - time initial belongs in the denominator if you're are finding Vavg.
Change in position i.e. y final - y initial belongs in the numerator.
ok

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