# Kinematics/Projectile Motion Question?

1. Feb 19, 2013

### Stanc

1. The problem statement, all variables and given/known data
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?

3. 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 time

Is this correct? or is there a step missing or am I going in the wrong direction?

2. Feb 19, 2013

### Staff: Mentor

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).

3. Feb 19, 2013

### Stanc

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?

4. Feb 19, 2013

### Staff: Mentor

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.

5. Feb 19, 2013

### Stanc

I think we are supposed to neglect air resistance

6. Feb 19, 2013

### Staff: Mentor

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