A rock is tossed straight up with a velocity of 37.1 m/s. When it returns, it falls into a hole 18.8 m deep.

How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

hfinal=hinitial+vi*time -1/2 9.8 time^2

hfinal is -18.8 hinital is zero.

solve for time, notice it is a quadratic.

To find the time the rock is in the air, we need to determine the time it takes for the rock to reach its highest point (when its velocity becomes zero) and then the time it takes for it to fall back down into the hole.

First, we can use the initial velocity of the rock (37.1 m/s) and the acceleration due to gravity (-9.8 m/s²) to find the time it takes for the rock to reach its highest point. We can use the kinematic equation:

v = u + at

where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (37.1 m/s)
a = acceleration (-9.8 m/s²)
t = time

Rearranging the equation to solve for time (t), we get:

t = (v - u) / a

Substituting the values, we have:

t = (0 m/s - 37.1 m/s) / (-9.8 m/s²)

Calculating this, we find:

t ≈ 3.78 seconds

The rock takes approximately 3.78 seconds to reach its highest point.

Next, we need to find the time it takes for the rock to fall back down into the hole. We can use the kinematic equation:

s = ut + (1/2)at²

where:
s = distance (18.8 m)
u = initial velocity (0 m/s as it starts from rest at the highest point)
a = acceleration due to gravity (-9.8 m/s²)
t = time

Rearranging the equation to solve for time (t), we get a quadratic equation:

0.5at² + ut - s = 0

Substituting the values, the equation becomes:

0.5(-9.8 m/s²)t² + 0t - 18.8 m = 0

We can solve this quadratic equation using the quadratic formula:

t = (-b ± √(b² - 4ac)) / (2a)

where:
a = 0.5(-9.8 m/s²) = -4.9 m/s²
b = 0 m/s
c = -18.8 m

Substituting the values, we get:

t = (± √((-0)² - 4(-4.9)(-18.8))) / (2(-4.9))

Simplifying further:

t = (± √(0 - 367.84)) / (-9.8)

Since we are looking for the positive time, we can disregard the negative root. Calculating this, we find:

t ≈ 1.98 seconds

Therefore, the rock takes approximately 1.98 seconds to fall back down into the hole.

To find the total time in the air, we sum the time for the ascent and the time for the descent:

Total time = time for ascent + time for descent
Total time = 3.78 seconds + 1.98 seconds
Total time ≈ 5.76 seconds

Therefore, the rock is in the air for approximately 5.76 seconds from the instant it is released until it hits the bottom of the hole.