I have a quadratic function word problem that I need help with. I need help with the questions about it. This is the word problem ------ "Suppose a baseball is shot up from the ground straight up with an initial velocity of 32 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."
These are the questions.
What function are they using in this problem?
The red words above answers that question.
The ball will be how high above the ground after 1 second?
Substitute v0 = 32, s0 = 0, and t = 1 in
s = -16t² + v0t + s0
s = -16(1)² + 32(1) + 0
s = -16(1) + 32
s = -16 + 32
s = 16 feet
How long will it take to hit the ground?
Substitute v0 = 32, s0 = 0, and s = 0 in
s = -16t² + v0t + s0
0 = -16t² + 32t + 0
16t² - 32t = 0
Divide through by 16
t² - 2t = 0
Factor out t on the left:
t(t - 2) = 0
Set each factor equal to 0:
t = 0, t - 2 = 0
t = 0, t = 2
The first solution tells us
the ball was on the ground at
the start. We already knew
that. The second solution tells
us that the ball is back on the
ground 2 seconds later. That is
the solution we want.
What time will the maximum height be attained?
The ball takes exactly the same amount
of time to go up as it takes to come
down, so since it takes 2 seconds to go
up and come down, it takes half that, or
1 second to go up, and 1 second to come
down.
What is the maximum height of the ball?
Substitute v0 = 32, s0 = 0, and t = 1 in
s = -16t² + v0t + s0
We already did that in the first part and
got 16 feet.
Edwin