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
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
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?
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?