SOLUTION: a) 2) Suppose you throw a baseball straight up at a velocity of 32 feet per second. A function can be created by expressing distance above the ground, s, as a function of time, t

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Question 36560: a)
2) Suppose you throw a baseball straight up at a 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:
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?
Answer:
Show work in this space.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
2) Suppose you throw a baseball straight up at a 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:
s(t)= -16t^2+32t
b) The ball will be how high above the ground after 1 second?
Answer:
Show work in this space.
s(1)=-16+32=16 ft. above the ground
c) How long will it take to hit the ground?
"Hit the ground" means the distance above the ground is "0".
0=-16t^2+32t
0=-16t(t-2)
t=0 and t=2
The ball will be on the ground at time t=0 and time t=2 sec.

d) What is the maximum height of the ball?
Answer:
Show work in this space.
The maximum occurs when t=-b/2a=-32/[(2)(-16)]= 1 sec.
Cheers,
Stan H.