
04/21/2005, 06:15 PM
#5276
Originally Posted by ChickDance
The height of an object that has been fired upward from the ground is found by using the formula below, where 'h' is the height in feet, 't' is the time in seconds since the object was fired, and 'v0' is the initial velocity with which it was fired, in feet per second:
h = v0t  16t2
A model rocket is fired upward at an initial velocity of 192 feet per second. After how many seconds is the rocket at a height of 432 feet above the ground?
Extra: What is the maximum height that the rocket reaches on its flight?
OK, lets see...
h = v0t  16t2 is a quadratic equation, there is a formula for that and a quadratic equation solver here.
With h = 432 feet and v0 = 192 feet per second we get h = 432 = 192t  16t2, or
16 t2 + 192t  432 = 0
For the quadratic equation (and the solver) we take a = 16, b = 192, and c = 432.
Solving the equation (using the solver) gives two possible t's, t1 = 3 seconds and t2 = 9 seconds (the rocket flies up at t=3 seconds and is on the way down at t=9 seconds).
The maximum hight must be reached in between the two times (3 and 9 seconds) at t=6 seconds. The first equation then gives
h = 192*6  16*6^2 = 576 feet.
(you can actually just copy/paste the formula 192*6  16*6^2 into google and you get the answer  Google is cool! )
HOWEVER: There is NO way a model rocket can fly according to the formula h = v0t  16t2. Model rockets are propelled on their way up (against gravity) until the "fuel" goes out, then they are slowed down by gravity until they reach maximum height, and fall back down following gravity alone. So the way up is totally different from the way down... it must be a trick question!


