perform problem correctly? algebra?

Question

3) Suppose a baseball is shot up from the ground straight up with an initial velocity of """64 feet per second""".( that means each second the ball will go up in the air 64 feet)
A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s = -16t2 + v0t + s0
•16 represents 1/2g, the gravitational pull due to gravity (measured in feet per second2).
•v0 is the initial velocity (how hard do you throw the object, measured in feet per second).
•s0 is the initial distance above ground (in feet). If you are standing on the ground, then s0 = 0.


What is the maximum height of the ball? What time will the maximum height be attained?
Answer: 128 feet., 2 seconds

Show work in this space.
s = -16t^2 + 64t + 0
max height happens halfway between the zeros of the quadratic, at t = 2. when t = 2, s = -16(4) + 2(64) = 64 ft.

s(2) = -16(2^2) + 64(2) = 128

Answer 1

Hi

Use the other equation

V = U + f t

Where V = final velocity
U = initial velocity
and t = time

Now 0 = 64 + (-32) t
f = (-32) because here it is against the gravity

0 = 64 - 32 t
32 t = 64
t = 2 Sec.

That means in 2 sec the velocity will comes to zero and the ball will start to come down.

Now using your function

s = -16 t^2 + 64 t
When t = 2

s = -16x 2^2 + 64 x 2
= -16x4 + 64 x 2
= - 64 + 128
= 64.

Then the maximum height reached by the ball is 64 feet and the time is 2 seconds.

Answer 2

s(2) = -16(2^2) +64(2) = 64 and not 128 as you say in you r Answer and in the final line. But in the 2nd to last sentence you correctly say max height = 64 feet.