SOLUTION: 3) 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,

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 3) 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,       Log On


   

Question 53230: 3) 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.
a) What is the function that describes this problem?
Answer:



b) The ball will be how high above the ground after 1 second?
Answer:
Show work in this space.



c) How long will it take to hit the ground?
Answer:
Show work in this space.



d) What is the maximum height of the ball? What time will the maximum height be attained?
Answer:
Show work in this space.
Thank You
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
If you need a calculus way of solving this, let me know and I will edit this to include calculus.
a) v0=32; s0=0;
s=-16t%5E2%2Bv0t%2Bs0
s=-16t%5E2%2B32t%2B0
s=-16t%5E2%2B32t
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b)s=-16%281%29%5E2%2B32%281%29
s=-16%2B32
s=16 ft.
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c) The position s=0 when the ball hits the ground.
0=-16t%5E2%2B32t
0=-16t%28t-2%29
-16t=0
-16t/-16=0/-16
t=0s. This is the time we threw the ball.
t-2=0
t-2+2=0+2
t=2 s. This is the time that the ball hit the ground.
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d)This is a parabolic function and it's maximum value occurs at its vertex. There are a couple of ways to find the vertex. If your taking caluculus, this is where you would do something different than what I'm showing you now.
The x value of the vertex can be found by putting your function in standard form.
s=at%5E2%2Bbt%2Bc
t=-b%2F%282%2Aa%29
In your problem:
t=-16t%5E2%2B32t
a=-16, b=32
t=-%2832%29%2F%282%2A-16%29
t=%28-32%29%2F%28-32%29
t=1 s. (At 1 s. the ball is as high as it will go.)
Substitute that into your function to find out how high it will go.
s=-16%281%29%5E2%2B32%281%29
s=-16%281%29%2B32%281%29
s=-16+32
s=16 ft. That is how high the ball will go in a perfect physics universe.