Question
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).
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.
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.
Answers
bobpursley
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
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
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
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
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