- #1

Terraist

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**1. A parachustist jumps from an airplane and freely falls y=49.4 m before opening his parachute. Thereafter, he decelerates at a=2.02 m/s2. As he reaches the ground, his speed is 3.11 m/s. How long was the parachutist in the air?**

For this one I managed to get the solution already using a method someone else showed me. However, I fumbled a long time with this problem and got the wrong answers despite having a strategy in mind.

What I did was the following. I split the problem up into two parts: before the parachute opens and after. These are the variables as I defined them at the beginning of the problem:

v

_{0}=0, s

_{0}= 0, for when parachutist jumps out plane.

v

_{1}= ?, s

_{1}=49.4m, a

_{1}= 9.8 m/s

^{2}, t

_{1}= ?, upon parachute release

v

_{2}= 3.11 m/s, s

_{2}= ?, a

_{2}= -2.02 m/s

^{2}, t

_{2}= ?, upon reaching ground

1. First I tried to find the time t

_{2}between parachute release and reaching ground.

0.5at

^{2}+ v

_{2}t + s

_{1}

0.5(-2.02 m/s

^{2})t

^{2}+(3.11 m/s)t +49.4m

using the Q-formula I got t

_{2}=8.712. Then I tried finding v

_{1}

v

_{2}=v

_{1}+at

_{2}

3.11 m/s = v

_{1}+ (-2.02 m/s

^{2})(8.71s)

v

_{1}= 20.7 m/s

3. Using v

_{1}, I could then find t

_{1}

t

_{1}= (v

_{1}-v

_{0})/a

t = 2.11s

The red part I think is where the mistake was. For that formula, I had to use v

_{1}, which I didn't know, but even if I did, would this approach still work? Plugging in values for v

_{1}, which I found later using a different method, still gave me wrong figures for t.

Rather than telling me how I could obtain a solution, please show me the mistake with this strategy and whether I could have tweaked it to make it work. I have struggled a long time with kinematics concepts and I admit my problem solving skills are subpar, so a physical description of what is occurring would be very helpful.

Thanks!