(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The height of an object moving vertically is given by

s = -16t^{2}+ 96t + 112,

whensin ft andtin 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

3. 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 = -16t^{2}+ 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 = -16t^{2}+ 96t + 112

0 = -16(t^{2}-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?

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# Homework Help: Applications of Derivative

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