SOLUTION: A ball is thrown vertically upward. After t seconds, its height
h (in feet) is given by the function h(t)=48t-16t^2
After how long will it reach its maximum height?
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: A ball is thrown vertically upward. After t seconds, its height
h (in feet) is given by the function h(t)=48t-16t^2
After how long will it reach its maximum height?
Log On
Question 1101446: A ball is thrown vertically upward. After t seconds, its height
h (in feet) is given by the function h(t)=48t-16t^2
After how long will it reach its maximum height? Found 3 solutions by htmentor, Alan3354, ikleyn:Answer by htmentor(1343) (Show Source):
You can put this solution on YOUR website! The maximum height will be reached when dh/dt = 0
dh/dt = 48 - 32t = 0
This gives t = 1.5
Ans: 1.5 seconds
You can put this solution on YOUR website! A ball is thrown vertically upward. After t seconds, its height
h (in feet) is given by the function h(t)=48t-16t^2
After how long will it reach its maximum height?
----------------
Without calculus:
The max is the vertex of the parabola.
That's at t = -b/2a = -48/-32 = 1.5 seconds
This question asks at which value of "t" the quadratic function
h(t) = 48 - 16*t^2
reaches its maximum.
The general theory regarding a quadratic function of general form
y = ax^2 + bx + c
says that it reaches the maximum/minimum at x = .
So, in our case, the maximal height reaches at t = = 1.5 seconds.
The referred lessons are the part of this textbook under the topic "Finding minimum/maximum of quadratic functions"
and under the topic "Projectiles launched/thrown and moving vertically up and dawn".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
