A helicopter rises vertically with a constant upward acceleration of 0.40 m/s2 . As it passes an altitude of 20 m, a wrench slips out the door.

A) How soon does the wrench hit the ground?

B) At what speed does the wrench hit the ground?

h = 20 + .4 t - 4.9 t^2

solve for t when h = 0 just like the problem you posted below this one.
also the speed when it hits the ground is the same method.

To find the answers to these questions, we can use the equations of motion under constant acceleration. The two equations that we'll be using are as follows:

1. v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
2. s = ut + 0.5at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

A) To find how soon the wrench hits the ground, we need to find the time it takes for the wrench to fall from an altitude of 20 m.

Given:
Initial velocity of the wrench (u) = 0 m/s (since the wrench fell freely from the helicopter)
Acceleration of the wrench (a) = acceleration due to gravity = 9.8 m/s^2 (assuming no air resistance)
Displacement of the wrench (s) = -20 m (negative because it's falling downward)

Using the equation s = ut + 0.5at^2, we can solve for t. Plugging in the given values:

-20 = 0 * t + 0.5 * 9.8 * t^2

Simplifying the equation:
-20 = 4.9 * t^2

Dividing by 4.9:
t^2 = -20 / 4.9

Taking the square root of both sides:
t ≈ 2.02 seconds (rounded to two decimal places)

Therefore, the wrench hits the ground approximately 2.02 seconds after it slips out of the helicopter.

B) To find the speed at which the wrench hits the ground, we need to calculate its final velocity (v) when it reaches the ground.

Given:
Initial velocity of the wrench (u) = 0 m/s
Acceleration of the wrench (a) = 9.8 m/s^2 (acceleration due to gravity)
Time taken (t) = 2.02 seconds (calculated in part A)

Using the equation v = u + at, we can calculate the final velocity (v). Plugging in the given values:

v = 0 + 9.8 * 2.02

Simplifying the equation:
v ≈ 19.8 m/s (rounded to one decimal place)

Therefore, the wrench hits the ground at approximately 19.8 m/s.