Question 71221: I have been trying to figure this out for two days and I am stuck. I have been through my text book and web site but I just don;t get it, can you help?
the question is not from a text book
thnks
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
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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.
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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.
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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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