An object is shot up into the sky according to s(t)= -16t^2+64t where s(t) = height above the ground in feet at t seconds. Find the a. velocity at 2 seconds b. acceleration at 1 second c. maximum height of the object.

Answer 1

v(t) = -32t+64
a(t) = -32
as with all quadratics, the vertex is at x = -b/2a

Answer 2

s(t)= -16t^2+64t
v(t) = -32t + 64
a(t) = -32

a) find v(2)
b) clearly a constant for the domain of t
c) set v(t) = 0 and solve for t
then sub that into s(t) to find the corresponding height

Answer 3

Let me see, I used feet a few years ago, like 60.
v = ds/dt = -32 t + 64
a = -32 ft/second^2 = constant acceleration of gravity by the way

max height when v = 0 (stops going up, not going down yet)
t = 64/32 seconds

Answer 4

Sorry, I'm not understanding c.

Answer 5

They asked for the maximum height
That is when ds/dt = 0
In other words when the speed up stops

Answer 6

So I'm plugging in 64 (which is what I got from putting 0 into v(t) ) into s(t) and my answer should come out to -61440? I'm confused lol

Answer 7

-b/2a = 64/-32 = -2
So find s(t) when t=2

You don't need any calculus for this. Just think back on your Algebra I
The vertex occurs midway between the roots, which are at t=0 and 4

Answer 8

v = ds/dt = -32 t + 64
0 = -32 t + 64
t = 2 not 64