Question 51276: I am unsure if I am doing this problem right.
Suppose a baseball is shot up from the ground straight up with an initial velocity of 32 feet per second. 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 by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! s = -16t2 + v0t + s0
s(t)=-16t^2+32t+0
Maximum at x=-b/2a
-b/2a=-32/(-32)= 1,
f(1)=-16(1)^2+32(1)=16
Maximum height is 16 ft at time = 1 second.
Cheers,
Stan H.
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