Expert: Ahmed Salami - 7/15/2010

Question
QUESTION: I know that v = d/t but I'm not sure how to set this up:The height of an object shot straight upward from the ground withan internal velocity of 64ft per sec is given by the equation h = 64t-16t ^2. It wants to know what the maximum height is that the object reaches. Could you please explain this to me? Thank you

ANSWER: Hi Cathy,
h = 64t - 16t²
We know that the object is at its maximum height when the velocity is zero (when it can no longer go up).
Now, velocity v is the rate of change of the height h
v = dh/dt
 = 64 - 32t
equating to zero,
64 - 32t = 0
t = 2 sec
This is the time it takes to get to the maximum height. The maximum height is
h = 64(2) - 16(2)²
 = 128 - 64
 = 64 ft

Regards

---------- FOLLOW-UP ----------

QUESTION: I just wondered where you got the 32t? I can see that you factored out a t, but wouldn't it still be 16t? Why is the 16 multiplied by 2?  Thank you for your help. I really appreciate it.

Answer
Hi Cathy,
I was of the opinion that you've done a bit of differential calculus.
v = dh/dt refers to instantaneous velocity which is quite different from h/t, average velocity.
You need to know how to differentiate a function to understand the idea of the question and even the solution.
As a general formula, for y = ax^n
dy/dx = nax^(n-1)

Also, we could solve this particular problem using basic algebraic techniques and reasoning.
h = 64t - 16t²
completing squares in order to rewrite the equation,
h - 64 = 64t - 16t² - 64
h - 64 = -16(t-2)²
h = 64 - 16(t-2)²
In this form its easier to figure out the maximum value of h.
Now, 16(t-2)² is always positive since the square of any number is positive. Since we're always subtracting this (positive) number from 64 to get h, it therefore means that h cant have a value higher than 64 which is then its maximum value. Note that this occurs when (t-2)² = 0 i.e t = 2 as we already got by differentiation.

Regards