SOLUTION: my homework question reads: an object is thrown upward from the top of a 256ft tower at a velocity of 96ft/sec. Find the height after 3 seconds. I know I need to use the quadra

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: my homework question reads: an object is thrown upward from the top of a 256ft tower at a velocity of 96ft/sec. Find the height after 3 seconds. I know I need to use the quadra      Log On


   

Question 571016: my homework question reads:
an object is thrown upward from the top of a 256ft tower at a velocity of 96ft/sec. Find the height after 3 seconds.
I know I need to use the quadratic equation, but I don't know what to do next.
3t=-16t^2+96t+256
16t^2-93t-256=0
Thanks for your help!
Answer by KMST(5398) About Me  (Show Source):
You can put this solution on YOUR website!
According to Physics, the height as a function of time in seconds should be
h%28t%29=-16t%5E2%2B96t%2B256
After 3 seconds, the object's height is
h%283%29=-16%2A3%5E2%2B96%2A3%2B256
h%283%29=-16%2A9%2B288%2B256
h%283%29=-144%2B288%2B256
h%283%29=400
The quadratic function h%28t%29 has a maximum and its graph (h versus t) is a parabola.
To figure out when it reaches its highest altitude, find the axis of symmetry of the parabola:
x=-288%2F%282%2A%28-16%29%29 --> x=9
So the object reaches its maximum height at 9 seconds.
That height is h%289%29=-16%2A9%5E2%2B96%2A9%2B256
To find out when it hits the ground (h%28t%29=0), you solve the quadratic equation
-16t%5E2%2B96t%2B256=0 or the equivalent 16t%5E2-96t-256=0