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.

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 ..

We take t as a common factor

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

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

Complete the square:

That is, has a maximum value of ft when 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.

if you was gonna solve for a, a is a   < -12

he kicked it up? is this the answer ?

step-by-step explanation: