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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:

2007-02-20 00:04:49 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

derivative s'=-32t+v0=-32t+64
s'=0 then t= 2(s' is the velocity function)
wich means at t=2 the baseball speed will be 0
s'' = -32
s'' has a negative value wich means that this system is uniformly decelerated , its velocity is decreasing with time.it decreased from 64 feet/s to 0 feet/s in 2 seconds.
if s'' is equal to 0 then the system is uniform.so the velocity will remain constant and will not change.
if s'' has a positive value then the velocity is increasing with time wich means the system is uniformly accelerated.
good luck

2007-02-20 00:24:44 · answer #1 · answered by babilla1983 1 · 0 0

taking upwards motion as positive....

s-s0=1/2 at^2 + v0t

a = -g = -32 ft/s^2

therefore....

s-s0 =1/2(-32)t^2 + v0t

s = -16t^2 + v0t +s0

2007-02-20 00:13:11 · answer #2 · answered by Anonymous · 0 0

Since we are throwing from the ground, we choose an axis such that s0 = 0 and the upward direction is positive.

s = ut + 0.5gt*t
s = 64t - 16t*t = 16t*(t - 4)

You actually have given the answer yourself. g is negative since it acts in the negative direction of choice of the axis.

2007-02-20 00:10:22 · answer #3 · answered by Shashi 2 · 0 0

You've told us what the function is, then asked us what it is.

The function is s = -16t² + v0t + s0

2007-02-20 00:10:00 · answer #4 · answered by Gnomon 6 · 0 0

Compound angles mixed with multifaceted surfaces, jumbled in with some various transitioning radius's. and then attempting to translate that right into a three-D solid CAD form. supplies me a huge headache, besides the undeniable fact that this is a great larger sense of alleviation whilst i eventually remedy it. :)

2016-10-02 10:46:49 · answer #5 · answered by ? 4 · 0 0

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