An object of mass 100g is thrown vertically upward from a point 60cm above the earths surface with an initial velocity of 150cm/sec. It rises briefly and then falls vertically to the earth, all of which time it is acted on by air resistance that is numerically equal to 200v (in dynes), where v is the velocity (in cm/sec).

Question

a) find the velocity 0.1sec after the object is thrown.

b)find the velocity 0.1sec after the object stops rising and start falling.

Answer 1

net force=mass*acceleration
mg-200v=m*a
a=g-200v/mass

note we have a sign issue. if g is always downward, then if v is upowrd, then
a=g-200v/m
but when the motion is downward,
a=g+200v/m because v is now a negative sign....so on part b), one has to change the sign..

a) now we have an issue with calculus. One is tempted to use
vf=vi+at, however, a should be average acceleration, not the acceleration at t. However, because we are using very small time incrments, one can approximate the average a as the a at final time (similar to Simpson's right hand rule on incremental integration).
So as an approximation
vf=vi+at=150-(98+200v/100 )* .1
v=150-(98+.2v)
v(1.2)=52 calculate v

b. Same technique, however be mindful fo the sign for friction.

v(downward)=0-980-200v/100).1
v==98+.2v
v=-98/1.2 cm/sec

Answer 2

got this when i solve a)
ma = F
m dv/dt = -mg -kv

Putting in known quantities,

100 dv/dt = -100(980) - 200v
dv/dt = -980 - 2v
dv/dt + 2v = -980
v(0) = 150

Answer 3

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