Thread: average velocity!!

If a ball is thrown into the air with avelocity of 80ft/s, its height in feet after t seconds is given by y= 80t-16t^2 . find its averge velocity from the time its thrown to the time it reaches its maximan height.
here is what i did
first i took the dervative of the function and equalled it to zero. i got t= 2.5.
then i calculated the average velocity i got 32 where it should be 40. i guess that i made a mistake while calculating the average velocity. can someone help me from there.
thanks in advance!!

From beginning: Derivative of the function is

Maximum height is: . The average velocity from beginning . Average velocity is:

So in t_0 = 2,5, the average velocity is:

Originally Posted by lukaszh

From beginning: Derivative of the function is

Maximum height is: . The average velocity from beginning . Average velocity is:

So in t_0 = 2,5, the average velocity is:

thanks but when u slove for the average velocity don't u take the initail velcity then subtract it from the final then divide it over the time which. that would be (80-0)/2.5 which makes it 32 ( atleast that we took in school) can u explain more to me plz?!!

Try to realise this situation. You product 300 toys per day. What's the average (in hours)? that is 300/24 = 12.5. So you procuct 12.5 toys per hour. Now I have a function, which is modelling the height in particular time. So the average is:

Numerator : Height
Denominator : quantity (time)

Originally Posted by lukaszh

Try to realise this situation. You product 300 toys per day. What's the average (in hours)? that is 300/24 = 12.5. So you procuct 12.5 toys per hour. Now I have a function, which is modelling the height in particular time. So the average is:

Numerator : Height
Denominator : quantity (time)

thanks. now i get it so if we substitute 2.5 in the main equation that will give is the hight which is 100 then divide it by the time 2.5 which gives us 40!!!!
thanks sooo much.