SOLUTION: 3) Suppose a baseball is shot up from the ground straight up with an initial velocity of 64 feet per second. A function can be created by expressing distance above the ground, s,

Question 71235: 3) Suppose a baseball is shot up from the ground straight up with an initial velocity of 64 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.

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 64 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:
s(t) = -16t^2 + 64t + 0
-------------
b) The ball will be how high above the ground after 1 second?
Answer:
Show work in this space.
s(1) = -16(1)^2 + 64(t)
s(1) = -16 + 64
s(1) = 48 ft.
-----------------------
c) How long will it take to hit the ground?
Answer:
Show work in this space.
Its height will be zero when it hits the ground.
So 0 = -16t^2 + 64t
t(-16t+64)=0
t= 0 or -16t+64=0
t=0 seconds or t= 4seconds
The height will be zero at the start and after 4 seconds.
----------------------------
d) What is the maximum height of the ball? What time will the maximum height be attained?
Answer:
Show work in this space.
s(t) = -16t^2 + 64t +0
This is a quadratic with a=-16, b=64, c=0
The high point is at t=-b/2a = -(64)/(-32) = 2 seconds
It reaches its highest point after 2 seconds.
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Cheers,
Stan H.

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