# Applications of Derivative

## Homework Statement

The height of an object moving vertically is given by
s = -16t2 + 96t + 112,
when s in ft and t in sec.
a. Find the object's velocity when t=0
b. Find its maximum height and when it occurs.
c. Find its velocity when s=0

## The Attempt at a Solution

a.
Now I know to find the objects velocity, take the derivative, which is:

v(t) = s' = -32t + 96.
when t=0, the objects velocity is 96 m/s

b.
This is when I got stuck. When the velocity is 96 m/s, what is its maximum height. I am guessing set the derivative's velocity equal to 0, and solve for t. and plug the value of t in the anti derivative equation, and you would get the maximum height, is the correct? Because whenever the derivative is zero, the anti derivative would either be a max or min, in this case, a max.

0 = -32t + 96
t = 3 sec

s = -16t2 + 96t + 112 plug t= 3 in this equation and you will get,
s = 256 ft?

c.

With the assumptions that my work is so far correct, this is what I would do. set the anti derivative equal to zero and solve for t, using factorization.

0 = -16t2 + 96t + 112
0 = -16(t2 -6t - 7)
0 = -16(t + 1)(t - 7)
t = -1, t = 7, because those are the values when the height is zero.

I would plug those values into the derivative to find the velocity at that point.

v(-1) = 128 m/s
v(7) = -128 m/s

Is all my work correct? Or did I made some mathematical or logical errors?

Last edited:

CompuChip
Homework Helper
Your answers are correct, except you used the wrong units in a).
Also in b, you say "When the velocity is 96 m/s, what is its maximum height.". That statement is nonsense, and you don't need it. The 96 m/s was about a specific time instance (t = 0), now we're talking about a different time instance (namely, t for which v is maximal) which you find by setting s' = 0.

Finally, I want to make a remark - perhaps a bit picky - about your use of the word "anti derivative". Personally, I would call s "the (distance / given / original) function", and v its derivative. If v was given, I would call that the given/original function and refer to s as the anti-derivative. Now when you say "anti-derivative" it may mean that you are looking at a function f, such that the derivative f' = s (and f'' = s' = v).

Your answers are correct, except you used the wrong units in a).
Also in b, you say "When the velocity is 96 m/s, what is its maximum height.". That statement is nonsense, and you don't need it. The 96 m/s was about a specific time instance (t = 0), now we're talking about a different time instance (namely, t for which v is maximal) which you find by setting s' = 0.

Finally, I want to make a remark - perhaps a bit picky - about your use of the word "anti derivative". Personally, I would call s "the (distance / given / original) function", and v its derivative. If v was given, I would call that the given/original function and refer to s as the anti-derivative. Now when you say "anti-derivative" it may mean that you are looking at a function f, such that the derivative f' = s (and f'' = s' = v).

Thank you so much.

And it was interesting how I never thought about the anti-derivative, like what it may mean when not used appropriately, thank you.