A mass of .25 kg is dropped from rest in a medium offering a resistance of .2abs

Question

A mass of .25 kg is dropped from rest in a medium offering a resistance of .2abs(v), where v is measured in m/s. a) if mass is dropped from a height of 30 m, find its velocity when it hits the ground

Explanation / Answer

With the resisting force acting opposite to the direction of motion, … my'' = mv' = ? mg + ? | v | --> v' = ? g + ( ? / m ) | v | --> v' = ? g + ? | v | …. where … ? = ? / m = 0.2 / 0.25 = 0.80 … Now, with the positive y axis upward and the body moving down … v = ( dy / dt ) < 0 … to make v > 0 … v --> ?v … but since | v | > 0 … … | v | = ?v ---> v' = ? g + ? | v | = ? g ? ? v = ? ? [ ( g / ? ) + v ] … v' = ? ? [ ß + v ] = dv / dt ---> dv / [ ß + v ] = ?? dt … where … ß = g / ? = 9.8 / 0.8 = 12.25… integrating … ln [ ß + v ] = ?? t + c1 … Taking the exponential … ß + v = exp [ ?? t + c1 ] = c1 exp [ ?? t ] … since exp [ c1 ] is also a constant … so that … v (t) = ?ß + c1 exp [ ?? t ] … starting from rest …v(0) = 0 ---> ?ß + c1 exp [ 0 ] = ?ß + c1 = 0 ---> c1 = ß … so that … v(t) = ?ß + ß exp [ ?? t ] = dy / dt ---> dy = { ?ß + ß exp [ ?? t ] } dt …integrating …y = ?ßt ? ( ß / ? ) exp [ ?? t ] + c2 = y(t) … where … ß / ? = ( g / ? ) / ? = g / ? ² … … ß / ? = 9.8 / ( 0.80 ) ² = 15.3125 … and … y(0) = 30 … so that … … 30 = 0 ? ( 15.3125 ) exp [ 0 ] + c2 = ?( 15.3125 ) + c2 … or … c2 = 30 + 15.3125 That is ... y(t) = ?( 12.25 ) t ? ( 15.3 ) exp [ ?( 0.80 ) t ] + 45.3 ... hitting ground means … y(t1) = 0 ---> ?( 12.25 ) t1 ? ( 15.3 ) exp [ ?( 0.80 ) t1 ] + 45.3 = 0 … solve for t1 … Let f(t) = ?( 12.25 ) t ? ( 15.3 ) exp [ ?( 0.80 ) t ] + 45.3 … where … … f(3) = ?( 12.25 ) (3) ? ( 15.3 ) exp [ ?( 0.80 ) (3) ] + 45.3 = 7.1620 > 0 … … f(4) = ?( 12.25 ) (4) ? ( 15.3 ) exp [ ?( 0.80 ) (4) ] + 45.3 = ?4.3237 < 0 … That means . 3 t2 = 1.69445 / 0.80 = 2.12 … … y(2.12) = ?( 12.25 ) ( 2.12 ) ? ( 15.3 ) exp [ ?( 0.80 ) ( 2.12 ) ] + 45.3 =16.5 … Maximum height of 16.5 m if no more than 10 m/s is to be attained.

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