A man jumps out of an airplane at an altitude of 16,000 feet and falls freely fo

Question

A man jumps out of an airplane at an altitude of 16,000 feet and falls freely for 20 seconds before opening his parachute. Assume linear air resistance, Fa = -kv. Assume that k = 0.18 without the parachute and k = 1.8 with the parachute.

Write down the differential equation that describes the velocity  before he opens hisparachute. What is his terminal velocity?

What will his velocity be at the instant he opens his parachute.

What will the mans altitude be at the moment he opens his parachute?

Write down the differential equation describing his motion after opening his parachute. Solve the IVP.

How long will it take for the man to finally reach the ground?

Explanation / Answer

1) before opening the parachute the ODE CAN be given as

dv/dt = -32.14 -0.18 v

and v(0) = 0

2) solving we get,

v(t) = 178.55 ( e-0.18t -1)

velocity when he opens the parachute

v(20) =178.55 ( e-0.2*20 -1)

=173.67 ft/s

3) height at which he opes parachute

is by intergrating

v(t) = 178.55 ( e-0.18t -1)

h(t) = -991.94 e-0.18t -178.55t +c

now use boundary condition

h(0) = 16000

so, c= 16991.94

now he opens parachute at

h(20) = 13393.83 ft

4) after opening the parachute

dv/dt = -32.14 -1.8 v

v(0) =173.67 ft/s

solving we get

v(t) = 17.78 e-1.8t -155.89

integrating we get

h(t) = 9.87 e-1.8t -155.89t +c

y(0) =13393.83 ft

so,

h(t) = 9.87 e-1.8t -155.89t +13383.96

5) now to find the time we solve the last equation

where h=0

t= 858.5 sec

so total time = 858.5+20=878.5 secs

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