A juggler tosses a ring into the air. The height of the ring in feet above the jugglers hands can be modeled by the function F(x) = -16xsquared + 16x., Where x is the time in seconds after the ring is tossed. Find the rings maximum height above the jugglers hands and the time it takes the ring to reach this height. Then find how long the ring is in the air.

Question

amistre64 · Answer

x is time since t for time would be to sensible i assume

amistre64 · Answer

-b/2a = max height for the time given and twice that is how long its in the air

amistre64 · Answer

i think across is frozed

across · Answer

o.o what in the world just happened

across · Answer

Amistre is right, but I will be a bit more wordy (for clarity's sake). ;P

You are told that the equation$$f(t)=-16t^2+16t$$models the trajectory of an object in the air. Let us find the maximum height this object attains. First, we need to differentiate this function:$$f'(t)=-32t+16.$$Then we set it equal to $0$ and solve for $t$:$$0=-32t+16\implies t=\frac{1}{2}.$$That is how many seconds it takes for the object to reach its maximum height. To determine its height at this time, it is a simple matter of plugging in a value:$$f(1/2)=-16(1/2)^2+16(1/2)\implies f(1/2)=4.$$That is how high the object travels (in whatever units you choose: you did not specify!). To determine how long the ring is in the air, find the roots of the function:$$-4t(4t-4)=0.$$Clearly, $t=0$ is a solution as is $t=1$. Therefore, the object stays in the air for one second.