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A find the velocity v t of the body at any time

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(a) Find the velocity v ( t ) of the body at any time before it is on the way down. (b) Use the result of part (a) to calculate the limit of v ( t ) as k 0 , that is, as the resistance approaches zero. Does this result agree with the velocity of a mass m projected upward with an initial velocity v 0 in a vacuum? (c) Use the result of part (a) to calculate the limit of v ( t ) as m 0 , that is, as the mass approaches zero. (a) Clearly, the velocity v ( t ) is given by (9), that is, v ( t ) = e - kt/m ( mg + kv 0 ) - mg k = v 0 e - kt/m + e - kt/m - 1 k mg. (b) When k 0 , on using the fact e x - 1 x as x 0, we obtain v ( t ) v 0 - gt. Thus, this result of course agree with the case that medium resistance could be negalected. (c) When m 0, we have e - kt/m 0 for t > 0, hence v ( t ) 0 . 2 Example 4. A ball with mass m is thrown upward with initial velocity 4

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v 0 from the roof of a building of height h. There is a force due to air resis- tance of k | v | . (a) Find the maximum height above the ground that the ball reaches. (b) (optional) Assuming that the ball misses the building on the way down, how to find the time that the ball hits the ground? (a) Recall (10) we know that the ball reaches the maximum height at T = m k log(1 + kv 0 mg ) . Then the maximum height above the ground reached by the ball equals h max = h + Z T 0 v ( t ) dt. Note that by (8) we have for t < T , m dv dt = - mg - kv. Thus on integrating both sides from 0 to T , and notice that v ( T ) = 0, we have - mv 0 = - mgT - k Z T 0 v ( t ) dt, thus Z T 0 v ( t ) dt = m k ( v 0 - gT ) = mv 0 k - m 2 g k 2 log(1 + kv 0 mg ) , therefore the height we want to find is given by h max = h + mv 0 k - m 2 g k 2 log(1 + kv 0 mg ) . (b) At T + t for t > 0, we have m dv dt = mg - kv and by (13) v ( T + t ) = mg (1 - e - kt/m ) k . Thus, mv ( T + t ) = mgt - k Z T + t T v ( s ) ds. 5
When the ball hits the ground, one has Z T + t T v ( s ) ds = h max , hence mv ( T + t ) = mgt - kh max .

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