In an action movie, the villain is rescued from the ocean by grabbing onto a ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase full of counterfeit money when he is 126.4 m above the water. If the briefcase hits the water 6.0 seconds later, how fast was the helicopter ascending?

To find the speed at which the helicopter is ascending, we can use the equation of motion:

s = ut + (1/2)at^2

Where:
s = displacement (vertical distance traveled by the briefcase) = 126.4 m
u = initial velocity (which is 0 since the briefcase is initially at rest)
t = time taken for the briefcase to hit the water = 6.0 s
a = acceleration (which is due to gravity and is approximated as -9.8 m/s^2, since it is acting in the opposite direction to the positive vertical axis)

Rearranging the equation, we have:

s = (1/2)at^2

Let's plug in the values:

126.4 = (1/2)(-9.8)(6.0)^2

Now, let's solve for a:

a = (2s) / (t^2)

a = (2 * 126.4) / (6.0)^2

a = 2.64 m/s^2

Since the acceleration due to gravity is acting in the opposite direction to the positive vertical axis, the acceleration of the helicopter must be equal in magnitude but opposite in direction. Therefore, the acceleration of the helicopter is 2.64 m/s^2 and is directed upwards.

To find the velocity (speed) at which the helicopter is ascending, we can use the equation:

v = u + at

Where:
v = final velocity (which is the speed of the ascending helicopter)
u = initial velocity (which is 0)
a = acceleration (which is 2.64 m/s^2)
t = time taken for the briefcase to hit the water (which is 6.0 s)

Plugging in the values, we get:

v = 0 + 2.64 * 6.0

v = 15.84 m/s

Therefore, the speed at which the helicopter is ascending is approximately 15.84 m/s.

To determine the speed at which the helicopter was ascending, we can use the concepts of kinematics and the equations of motion.

First, let's analyze the motion of the briefcase. We know that the only force acting on the briefcase is gravity, which causes it to accelerate at a rate of 9.8 m/s² (acceleration due to gravity). The distance it travels in the vertical direction is 126.4 m, and the time it takes to reach the water is 6.0 seconds.

Using the equation:

d = v₀t + (1/2)at²

where:
- d is the distance traveled
- v₀ is the initial velocity (which is 0, as the briefcase starts from rest)
- t is the time taken
- a is the acceleration

Rearranging the equation to solve for acceleration:

a = (2d - v₀t) / t²

Substituting the given values, we have:

a = (2 * 126.4 m - 0 m/s * 6.0 s) / (6.0 s)²
a = 253.6 m / (36 s²)
a = 7.044 m/s²

The briefcase experienced an acceleration of 7.044 m/s² due to gravity.

Now, since the helicopter is the reference point and the briefcase is falling, the acceleration of the briefcase due to gravity is also the acceleration of the helicopter.

Therefore, the speed at which the helicopter was ascending is 7.044 m/s.

If the briefcase hits the water at t = 6.0 s, and the water was 126.4 meters below,

V*t - (1/2)g*t^2 = -126.4 meters
6 V - (4.9)*36 = -126.4

Solve for V, in meters/second

6 V = 50
V = 8.33 m/s