(adsbygoogle = window.adsbygoogle || []).push({}); 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}=8.71

2. 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 occuring would be very helpful.

Thanks!

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# Homework Help: Two part 1d kinematics problem. Need help, not solution!

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