Question 85323: 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 ½g, the gravitational pull due to gravity (measured in feet per second 2).
• 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?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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 ½g, the gravitational pull due to gravity (measured in feet per second 2).
• 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.
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s(t) = -16t^2+32t+0
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What is the maximum height of the ball?
max height occurs at t = -b/2a = -32/(2(-16)) = 1
max height is s(1) = -16+32 = 16 feet
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Cheers,
Stan H.
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