Question 274047: Quadratic Functions:
"The height of a ball thrown straight upward is measured by the function h(t)=-16t^2+40t, where h(t) is measured in feet and t is time in seconds."
A) Evaluate h(1) and explain what the dependent and independent variables represent.
B) What is the height of the ball after 2.5 seconds?
C) When is the ball at a height of 48 ft.?
D) When will the ball hit the ground?
* I have answered the first part of (A), getting a solution of 24. I am not sure what the dependent and independent variables are.
*I have solved (B) for an answer of 0.
*For part (C), I am not sure how to go about solving this backwards.
*For part (D), I can't even begin to know what to do. How should I go about solving for this, do I even have enough information?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! The height of a ball thrown straight upward is measured by the function h(t)=-16t^2+40t,
where h(t) is measured in feet and t is time in seconds."
:
A) Evaluate h(1) and explain what the dependent and independent variables represent.
24 ft like you said, the dependent variable is h(t), so called because it's value
depends on the value of t, t is the independent variable
:
B) What is the height of the ball after 2.5 seconds?
h(2.5) = -16(2.5^2) + 40(2.5) = 0; (which you got)
:
C) When is the ball at a height of 48 ft.?
Substitute 48 for h(t), find t
-16t^2 + 40t = 48
-16t^2 + 40t - 48 = 0
Simplify, and change the signs divide by -8, (easier to factor)
2t^2 - 5t + 6 = 0
There are not any real solutions for this equation, means it never reaches 48 ft
the discriminant (25 - 48) is negative
:
D) When will the ball hit the ground?
you already found that out in (B) when the h(t) = 0 (ground level)
:
A graph of this equation will help here
Note that it never gets close to 48 ft, the highest point was 25 ft at 1.25 sec
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