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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 of time, t. This function is s=16t squard+v0t+so.
16 represents 1/g, the gravitational pull due to gravity (measured in feet per second squared.
v0 is the initial velocity (how hard do you throw the object, measured in feet per second squared).
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?
s(t)=-16t^2+64t+0
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b)The ball will be how high above the ground after 1 second?
s(1) = -16+64+0 = 48 ft above the ground
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c)How long will it take to hit the ground?
When it hits the ground its height is zero:
-16t^2+64t=0
-16t(t-4)=0
t=0 or t=4
This means it is on the ground when it starts and 4 seconds later.
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d)What is the maximum height of the ball? What time will the masimum height be attained?
maximum occurs at t=-b/2a = -64/(-32) = 2 seconds
s(2) = -16*(2)^2 +64*2
s(2) = -64 + 128
s(2) = 64 ft above the ground
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Cheers,
Stan H.
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