Question 81749: suppose a baseball is shot up from the ground straight up with an initial velocity of 64 feet per second. a function canbe created by expressing distance above the ground,s,as a function of time,t. this function is s=-16t^2+vot+so.
>16 represents 1/2g, the gravitational pull due to gravity(measured in feet per second^2)
>vo is the initial velcoity(how hard do you throw a object, measured in feet per second)
>So is the initial distance above ground(in feet). if you are standing on the ground, then So=0
-what is the function that describes this problem
-the ball will be how high above ground after 1 second
-how long will it take to hit ground
-what is the max height of the ball? what time will the max height be attained?
Answer by bucky(2189)   (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=-16t^2+vot+So
.
16 represents 1/2g, the gravitational pull due to gravity (measured in feet per second^2)
.
vo is the initial velocity (how hard do you throw a object, measured in feet per second)
.
So is the initial distance above ground (in feet). If you are standing on the ground,
then So=0
.
********************************
-What is the function that describes this problem?
.
All you have to do is to substitute the given values into the equation for S. Since the
ball is being shot upward with an initial velocity of 64 fps, vo = 64. And since it is being
launched from the ground, the initial height, So, is equal to zero. So the equation
becomes:
.
S = -16t^2 + 64t + 0
.
and you just drop the trailing 0 to make it:
.
S = -16t^2 + 64t
**************************
The ball will be how high above ground after 1 second?
.
Just use the equation for S and substitute 1 for t to get:
.
S = -16(1^2) + 64*1
.
This simplifies to:
.
S = -16*1 + 64 = -16 + 64 = 48
.
So 1 second after launch the ball is 48 feet above the ground.
**********************************
How long will it take to hit ground?
.
When it hits the ground, its height, S, will be zero. Therefore, set S equal to zero
and solve the equation for t. Setting the height equal to zero makes the equation:
.
0 = -16t^2 + 64t
.
If you divide both sides of this equation by -16 the equation is reduced to:
.
0 = t^2 - 4t
.
Factor the right side and the equation is converted to:
.
0 = t(t - 4)
.
Notice this equation will be true if either of the factors is equal to zero because
it will involve a multiplication by zero on the right side ... and zero times anything
is zero.
.
Therefore, either t = 0 or t - 4 = 0. But t = 0 is the time just at the instant of launch
and we know the ball is at ground level at that time. The next time the ball reaches
ground level is obtained from the factor t - 4. It will equal zero when t = +4. So 4 seconds
after the ball is launched it hits the ground.
.
*********************************
.
-what is the max height of the ball? what time will the max height be attained?
.
This problem is easier solved in reverse. The ball shoots up and returns to the ground
in 4 seconds. It spends the first 2 seconds rising and the second 2 seconds falling.
So it will reach its maximum height 2 seconds after launch.
.
The height it will reach can be obtained by substituting 2 seconds for t in the height
equation. If you do that substitution the height equation becomes:
.
S = -16(2^2) + 64*2
.
This simplifies to:
.
S = -16*4 + 64*2 = -64 + 128 = 64
.
The ball rises to a height of 64 feet at the peak of its flight.
.
Hope this helps you to understand the problem a little better. This type of problem is
a fairly common exercise when you study gravity in Physics.
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