A toy rocket is fired upward from the ground. The relation between its height h, in feet, and the time t from launch, in seconds, can be described by the equation:

h=-16r^2 + 64t. How long does the rocket stay more than 48 feet above the ground?

Question

A toy rocket is fired upward from the ground. The relation between its height h, in feet, and the time t from launch, in seconds, can be described by the equation:

h=-16r^2 + 64t. How long does the rocket stay more than 48 feet above the ground?

anonymous · Answer

Firstly, tell your physics teacher to use metric, it's a far superior system internationally and generally. Secondly, you want to find the two values of t whereby h = 48.

So, 48 = -16t^2 + 64t (Presuming your "r" was actually a "t")
3 = -t^2 + 4t
t^2 - 4t + 3 = 0
(t-3)(t-1) = 0
t = 3 or t = 1

So we can describe this, by saying that the rocket reaches a height of 48 foot at time = 1, and exceeds after that, before dropping down and reaching it for the second time at time = 3. Therefore the time it remains above is 3-1 = 2 seconds.

anonymous · Answer

Lol your funny,  but thank you anyway