classical physics 300 level, University level. hw2.pdf Adobe Reader 139% Problem

Question

classical physics 300 level, University level.

hw2.pdf Adobe Reader 139% Problem 7 (15 points): An object is released from rest far above the ground. The object experiences a force due to the air drag that is proportional to the object's speed f--bv. In this problem let's define the y-axis to point down. (a) Your friend, who is also taking Classical Mechanics has calculated >, (f). He/she knows that the terminal velocity of the object is 40 m/s, and has found a solution. v.. (t) = 40 ( l-e-t /2 ) , but isn't confiden that it isight. In physics we often want to check our answers by picking situations where you independently know what the answer should be without using your formula, and then checking that your proposed formula gives these answers. In this case, you know v.(t) for t = 0 and t x from your basic understanding of physics. Check your friend's solution at these two points. How is his/her equation looking so far? (b) Given that the drag force on this object is f the drag force is not very significant for small times. What is the formula for y, (r) in the case of no drag? (c) Given the above. check your friend's solution by finding an approximation that is valid for small t. If you find that it is incorrect, can you suggest to your friend what term(s) in the equation he she might want to double check?

Explanation / Answer

part a:

at t=0, e^(-t/2)=1

then vy(t)=40*(1-1)=0

at t=infinity, e^(-t/2)=0

then vy(t)=40*(1-0)=40 m/s

so when released , the velocity is 0 m/s and terminal velocity is 40 m/s.

hence the formula, based upon the initial and final condition is correct.

part b:

in case of no drag, acceleration due to gravity is the only acting force.

hence vy(t)=initial speed+acceleration*time

=0+9.8*t

=9.8*t

part c:

e^(-t/2)=1-(t/2)+(t/2)^2+….

excluding 2nd and higher order, e^(-t/2)=1-0.5*t

hence for small t, vy(t)=40*(1-(1-0.5*t))=20*t

so the friend has to check the exponent in e^(-t/2). the “2” seems to be wrong.

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