The perfect parachute

Classical Mechanics Level 5

A parachutist of mass m m jumps from height h h and opening her parachute after t t seconds. A retarding force given by F ⃗ = − k v ⃗ \vec { F } =-k\vec { v } acts on her so that she lands on the ground with no velocity. Find h − 1 2 g t 2 h-\frac { 1 }{ 2 } g{ t }^{ 2 } .

This problem is originally part of set Mechanics problems by Abhishek Sharma .

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m g t k + m 2 g k 2 ln ⁡ ( 1 + k t m ) \frac { mgt }{ k } +\frac { { m }^{ 2 }g }{ { k }^{ 2 } } \ln { (1+\frac { kt }{ m } ) } m g t k − m 2 g k 2 ln ⁡ ( 1 + k t m ) \frac { mgt }{ k } -\frac { { m }^{ 2 }g }{ { k }^{ 2 } } \ln { (1+\frac { kt }{ m } ) } This scenario is not possible. m g t k − m 2 g k 2 ln ⁡ ( 1 − k t m ) \frac { mgt }{ k } -\frac { { m }^{ 2 }g }{ { k }^{ 2 } } \ln { (1-\frac { kt }{ m } ) } m g t k + m 2 g k 2 ln ⁡ ( 1 − k t m ) \frac { mgt }{ k } +\frac { { m }^{ 2 }g }{ { k }^{ 2 } } \ln { (1-\frac { kt }{ m } ) }

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Abhishek Sharma by Abhishek Sharma , 22, India

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1 solution

Callie Ferguson
Nov 12, 2019

In order for this to be true, the value of k k must be zero. This is not possible because that would mean there is no retarding force.

m d v d t = − k ( v − m g k ) m\frac{dv}{dt} = -k(v-\frac{mg}{k})

→ m d v = − k ( v − m g k ) d t \rightarrow m dv = -k(v-\frac{mg}{k}) dt

Integrating the above equation gives:

v ( t ) = m g k ( 1 − e − k t ) v(t) = \frac{mg}{k}(1-e^{-kt})

The problem gives that the velocity is zero, so we need to set the above equation to zero and solve for t t . This gives…

1 = e − k t 1 = e^{-kt}

l n ( 1 ) = − k t = 0 ln(1) = -kt = 0

− k t = 0 -kt = 0

We know that t t is greater than zero because time has passed, so in order for the above equation ( − k t = 0 ) (-kt=0) to be true, the constant k k must be zero. Again, this is not possible because this would indicate that there is no retarding force.

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