Rectilinear Motion (Calculus) Problem: A body move vertically up from the earth according to $s= 64t-16t^2.$ Show that it has lost one-half its velocity in its first $48$ ft of rise.
My Answer: I really don't understand the problem and totally clueless on what to do but I try this:
$$V=64-16t$$
I let, $48=64t-16t^2,$ and I get $t=3.$
When $t=3,$
$$V= -32 \mbox{ ft./sec.}$$
But it's not $1/2.$ Please I really need to know this because tomorrow is the deadline of our homework and maybe I will be called for presentation of answer. This is the last problem on our homework set.
 A: As hinted in a comment, the equation for $V$ actually should be written
$$ V = 64 - 32t. $$
This is evidently the equation you actually used, since it agrees with the result that $V=-32$ when $t=3.$
In other respects the approach given in the question is (mostly) correct, 
but is missing one small fact that resolves the problem.
If you make a graph of the height of the object as a function of time, it will
be a parabola opening downward. The starting position ($x=0, t=0$) is on
the left-hand part of the parabola where $s$ is still increasing as $t$ increases.
Eventually $s$ reaches its maximum value $s_{\max}$, then decreases again.
So the object will pass through every $s$ between $0$ and  $s_{\max}$ twice,
once on the way "up" and once on the way "down".
This corresponds to the fact that an equation of the form
$ at^2 + bt + c = 0$
very often has two real solutions.
In your case you have correctly written the equation
$$ 48=64t-16t^2,$$
which is an $ at^2 + bt + c = 0$ equation (just slightly rearranged)
and which has two solutions, $t=1$ and $t=3.$
You have simply missed one of the solutions and gotten the "other" one.
You want the time when the object first passes through $s = 48$,
so of the two values of $t$ that solve your quadratic equation,
you want the lesser positive value, $t=1.$
A: $$
s=64t-16t^2 \rightarrow v(t) = \frac{ds}{dt} = 64 - 32 t
$$
$$
v(t=0) = 64
$$
$$
s = 48 = 64t - 16t^2 \rightarrow (t-3)(t-1) = 0
$$
The problem states "in its first 48 ft " so $t = 1$
$$
v(t=1) = 64 - 32 (1) = 32
$$
Clearly this shows that is lost one-half of its velocity.
