Question

The average cricket jumps vertically with an initial upward of 10 ft/s. What is the hang time of such a jump, ignoring air resistance? Use the formula h=-16t^2+10t, where h is the height of the cricket in feet and t is the time in seconds after the jump. Round your answer to the nearest tenth.

Answer 1

Complete the square:

$-16t^2+10t=-16\left(t^2-\dfrac58t\right)=-16\left(t^2-\dfrac58t+\dfrac{25}{256}-\dfrac{25}{256}\right)=-16\left(t-\dfrac5{16}\right)^2+\dfrac{25}{16}$

That is, $h(t)$ has a maximum value of $\dfrac{25}{16}\approx1.6$ ft when $t=\dfrac5{16}\approx0.3$ s. It takes the cricket twice as much time to jump up to its maximum height and return to the ground, so that the hang time is about 0.6 s.

Answer 2

Answer:

The hang time is 0.63 seconds

Step-by-step explanation:

We want to find out what the cricket's fall time is and we have the function that describes its height as a function of time.

The cricket is on the ground when its height is equal to zero. Therefore we must equal h to zero and solve the equation for t ..

$h=-16t^2+10t\\\\-16t^2 +10t=0$

We take t as a common factor

$-16t^2 +10t=0\\\\t(10-16t)=0$

Then the height is zero when t = 0 and when (10-16t) = 0

$t=0$

$10-16t= 0$

$10 = 16t$

$t=\frac{10}{16}\\\\t= 0.63\ sec$

At t = 0 seconds the cricket is still on the ground.

Then the cricket is in the air and after 0.63 seconds the cricket falls back to the ground