A baseball is thrown upward from a height of 2m with an initial velocity of 10m/s. Determine its maximum height.

S(0) = 2m
S'(0) = 10m/s

S'(t) = -32t + 10
S(t) = -16t^2 + 10t + 2

I know that if you equal s(t) = 0 you get time, but how do I get maximum height?

Thanx in advanced!

answer is 7.1m ... i just don't know how to get there

To find the maximum height of the baseball, you need to determine the time at which it reaches its peak. The maximum height occurs when the velocity of the baseball becomes zero.

Now, we can find the time it takes for the velocity to become zero by setting S'(t) = 0:

-32t + 10 = 0

To solve for t, we can isolate t:

-32t = -10
t = -10 / -32
t = 5/16 ≈ 0.3125 seconds

So, the time it takes for the velocity to become zero is approximately 0.3125 seconds.

To find the maximum height, substitute this time value (0.3125 seconds) into the position equation S(t):

S(t) = -16t^2 + 10t + 2

S(0.3125) = -16(0.3125)^2 + 10(0.3125) + 2
S(0.3125) ≈ 3.0469 meters

Therefore, the maximum height of the baseball is approximately 3.0469 meters.