Paleontologists estimate that if a Tyrannosaurus rex were to trip and fall, it would have experienced an upward normal force of approximately 261,100 N acting on its torso when it hit the ground. Assume the torso has a mass of 3700 kg.

Assuming the torso is in free fall for a distance 1.37 m as it falls to the ground, how much time is required for the torso to come to rest once it contacts the ground?

I have determined the upward acceleration to be 60.75m/s2.
I am having trouble dterminign what equation to use.

To determine the time required for the torso to come to rest once it contacts the ground, you can use the equation of motion:

v^2 = u^2 + 2as

Where:
- v is the final velocity (which is 0 m/s since the torso comes to rest)
- u is the initial velocity (which is the velocity just before the torso hits the ground, which is unknown)
- a is the acceleration (which is the upward acceleration, 60.75 m/s^2, in this case)
- s is the distance traveled (1.37 m, in this case)

Rearranging the equation, we have:

0 = u^2 + 2 * 60.75 * 1.37

Simplifying further, we get:

u^2 = - (2 * 60.75 * 1.37)

To solve for u, we take the square root of both sides:

u = √(- (2 * 60.75 * 1.37))

Now, we have the initial velocity just before the torso hits the ground. Since the torso is in free fall, the initial velocity would be the velocity at the moment it starts falling, which is 0 m/s. Therefore, we must consider the negative value when calculating the square root.

Since the square root yields a complex number in this case, it means that the given values are inconsistent. It is not physically possible for the torso to have an upward acceleration of 60.75 m/s^2 while falling freely.

Please double-check the given values or provide additional information if available to determine the correct equation and solution.