# Assistance with Vertical Terminal Velocity motion - What do you think?

1. Apr 3, 2008

### TheKovac

1. The problem statement, all variables and given/known data
A parachutist is falling vertically with a uniform speed of 12 m/s. At the instant he is 50m above the ground, he drops a coin. (Ignore Air Resistance)
A) What is the seed with which the coin strikes the ground?
b) How long does it take for the coin to reach the ground

C) WHAT IS THE TIME DIFFERENCE BETWEEN THE COIN AND PARACHUTIST REACHING THE GROUND?

2. Relevant equations
v^2 = u^@ + 2ax
x=Ut+0.5at^2
x=0.5(u+v)t
v=u+at

3. The attempt at a solution
a)
=> v^2 = 12^2 +2(9.8)(50)
final velocity of coin= 33.53m/s RIGHT

b)
x=ut + 0.5at^2
50= 12t +0.5(9.8)t^2
4.9t^2 +12t -50=0
time for coin to reach ground = 2.19s

c) =???

Is this question able to be solved, with the current data, because I believe I would need mass or total displacement to correctly work the answer.

Kindest Regards,
TheKovac

2. Apr 3, 2008

### Hootenanny

Staff Emeritus
HINT: The parachutist is traveling at constant velocity.

3. Apr 3, 2008

### TheKovac

My goodness, could be as simple as:

t= x/a
=> t= 50/9.8

Which would result in:

t= 4.16s
=> t(coin)= 2.16s
=> t(parachute) = 4.16s
t(difference) = 2s

Is that how simple the answer is?

4. Apr 3, 2008

### Hootenanny

Staff Emeritus
Your close, but be careful. What is the definition of velocity?

5. Apr 3, 2008

### TheKovac

Velocity = Vector quantity of the rate of change of displacement.

So I am missing direction? Thats it?

6. Apr 3, 2008

### Hootenanny

Staff Emeritus
Not quite, speed is defined as the magnitude of velocity, or the rate of change of distance with time,

$$|v| = \frac{dx}{dt}$$

not

$$|v| \neq \frac{da}{dt}$$

as you had. In this case the acceleration is zero.

7. Apr 3, 2008

### TheKovac

So with that in mind, is this how to work the difference?

If the velocity is constant, then can we use t = x/v ?

t(parachutist)= 50/12
t(parachutisit)= 4.16
t(coin) = 2.16s
t(difference) = 2s

I think this one is right?

I worked out that my problem above, is that I used the equation for acceleration, when Acceleration was Zero, instead I should have used the equation for velocity, right?

8. Apr 3, 2008

### Hootenanny

Staff Emeritus
I've not checked your arithmetic, but your method is correct, as it is for all the previous questions