Question

crickets can jump with a vertical velocity of up to 14 ft/s. which equation models the height of such a jump, in feet, after t seconds?
( h(t)=-16t^{2}+v_{0}t+h_{0} )
( h(t)=-16t^{2}+v_{0}t + 14 )
( h(t)=-16t^{2}+14t )
( h(t)=-16t^{2}+14t + 14 )
what is the maximum height the cricket reaches? round to the nearest thousandth.
( h=square ) feet
complete
what is the crickets hang time (amount of time the cricket is airborne)? round to the nearest thousandth.
( t=square ) seconds
done

Explanation:

Step1: Find the maximum height

The function for height is \(h(t)=-16t^{2}+14t\). This is a quadratic function in the form \(y = ax^{2}+bx + c\) (\(a=-16\), \(b = 14\), \(c = 0\)). The \(t\) - value of the vertex of a quadratic function \(y=ax^{2}+bx + c\) is given by \(t=-\frac{b}{2a}\).
Substitute \(a=-16\) and \(b = 14\) into the formula: \(t=-\frac{14}{2\times(-16)}=\frac{14}{32}=\frac{7}{16}\)
Then substitute \(t = \frac{7}{16}\) into the height function \(h(t)\):
\(h(\frac{7}{16})=-16\times(\frac{7}{16})^{2}+14\times\frac{7}{16}\)
\(=-16\times\frac{49}{256}+\frac{98}{16}\)
\(=-\frac{49}{16}+\frac{98}{16}=\frac{-49 + 98}{16}=\frac{49}{16}=3.0625\)

Step2: Find the hang - time

Set \(h(t)=0\), so \(-16t^{2}+14t=0\). Factor out a \(t\): \(t(-16t + 14)=0\)
We have two solutions: \(t = 0\) (corresponds to the time when the cricket starts the jump) and \(-16t+14=0\)
Solve \(-16t+14=0\) for \(t\):
\(-16t=-14\), then \(t=\frac{14}{16}=\frac{7}{8}=0.875\)

Answer:

The maximum height \(h = 3.0625\) feet. The hang - time \(t=0.875\) seconds.