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Question 79607: Does anyone have any idea what this problem means? Any help would be appreciated!
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.
a) What is the function that describes this problem?
Answer:
b) The ball will be how high above the ground after 1 second?
Answer:
c) How long will it take to hit the ground?
Answer:
d) What is the maximum height of the ball?
Answer:
Show work in this space.
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.
a) What is the function that describes this problem?
Answer:
Vo=32 ; So=0 because the ball started at a height of zero.
EQUATIOn:
h(t)=-16t^2+32t+0
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b) The ball will be how high above the ground after 1 second?
Answer:
h(1)=-16+32 = 16 ft (that is how high the ball will be after one second)
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c) How long will it take to hit the ground?
Answer:
The height of the ball will be zero when it is back on the ground:
0 = -16t^2+32t
-16t(t-2)=0
t=0 or t=2
The ball started on the ground and will be back at the end of 2 seconds.
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d) What is the maximum height of the ball?
Answer:
Show work in this space.
Max height occurs when t=-b/2a = -32/2*(-16) = 1
Max height occurs at the end of 1 second.
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Cheers,
Stan H.
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