Kinematics/Projectile Motion Question?

[Stanc]

Homework Statement

If one jump from an airplane 4572m above the ground. if the terminal speed of a human is 53m/s (without a parachute) how long would it take u to hit the ground and how fast would u be going when u hit the ground?

The Attempt at a Solution

I answered the second part with just 53m/s since its the terminal speed

For part 1 however, I first found the time it takes for the human to reach 53m/s soo:
9.81 = (53 - 0)/t... t = 5.4s...With that i found the distance it travels during this time period : d = (0)(5.4) + (1/2)(9.81)(5.4)^2... d = 143m

With this distance, I subtracted it from the original distance: 4572 - 143 = 4429m
With that distance, I found the time it takes so: t = 4429/53 = 83.56s... then I added the 2 times: so 83.56 + 5.4 = 88.96s = total timeIs this correct? or is there a step missing or am I going in the wrong direction?

 

[mfb]

A human does not accelerate with 9.81m/s^2 until it reaches its final velocity and stays at that velocity (with zero acceleration) afterwards. It should give a reasonable approximation, however. The real acceleration begins at 9.81m/s^2 and goes down as the speed gets closer and closer to the terminal speed (which is never reached in an ideal model). A better evaluation would need a differential equation (and its solution).

 

[Stanc]

A human does not accelerate with 9.81m/s^2 until it reaches its final velocity and stays at that velocity (with zero acceleration) afterwards. It should give a reasonable approximation, however. The real acceleration begins at 9.81m/s^2 and goes down as the speed gets closer and closer to the terminal speed (which is never reached in an ideal model). A better evaluation would need a differential equation (and its solution).

So would I do something like this:
d = (Vi)t +(1/2)(a)t^2
4572m = 0m + (0m/s)(t) + (1/2)(-9.81m/s^2)(t^2)
4572m = (-4.905m/s^2)(t^2)
-932.1s^2 = t^2
30.53s = t

Vf = Vi + at
Vf = 0m/s + (-9.81m/s^2)(30.53s)
Vf = -299.5m/s
But since you cannot exceed terminal you will still be going at 53m/s?

 

[mfb]

No, that would neglect air resistance completely...
Do you know how to calculate the force due to air drag? With F=m*a, this gives a velocity-dependent acceleration.

 

[Stanc]

I think we are supposed to neglect air resistance

 

[mfb]

The terminal speed is the result of air resistance (and gravity) only.