SOLUTION: April shoots an arrow upward into the air at a speed of 64 feet per second from a platform that is 34 feet high. The height of the arrow is given by the function h(t) = -16t2 + 64t

Question 1119700: April shoots an arrow upward into the air at a speed of 64 feet per second from a platform that is 34 feet high. The height of the arrow is given by the function h(t) = -16t2 + 64t + 34, where t is the time is seconds. What is the maximum height of the arrow?
Found 2 solutions by ikleyn, ankor@dixie-net.com:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

To solve the problem, find the maximum of the quadratic function h(t) = -16t^2 + 64t + 34. The maximum is at t= , where I refer to the general form of a quadratic function f(x) = ax^2 + bx + c. In your case, a = -16, b = 64, c = 34, so the time when h(t) gets maximum is t = = = 2 seconds. Substitute this value of t into h(t) = -16t^2 + 64t + 34 to find the maximal height at this moment.

To see other similar solved problems, look into the lessons
    - Problem on a projectile moving vertically up and down
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower
    - Problem on a toy rocket launched vertically up from a tall platform

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook 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

to your archive and use it when it is needed.

Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
April shoots an arrow upward into the air at a speed of 64 feet per second from a platform that is 34 feet high.
The height of the arrow is given by the function h(t) = -16t^2 + 64t + 34, where t is the time is seconds.
What is the maximum height of the arrow?
:
The max height occurs on the axis of symmetry: Using x = -b/(2a) where
a = -16; b = 64
x =
x = 2 seconds
Find the height at 2 sec
h(t) = -16(2^2) + 64(2) + 34
h(t) = -64 + 128 + 34
h(t) = 98 ft is max height
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