Question

This is a differential equations question. It is dropped from a height of 300m. The falling object in Example 2 satisfies the initial value problem dv/dt = 9.8 - (v/5), v(0) = 0. (a) Find the time that must elapse for the object to reach 98% of its limiting velocity. (b) How far does the object fall in the time found in part (a)? Modify example 2 so that the falling object experiences no air resistance. (a) Write down the modified initial value problem. (b) Determine how long it takes the object to reach the ground. (c) Determine its velocity at the time of impact.

          This is a differential equations question. It is dropped from a height of 300m. The falling object in Example 2 satisfies the initial value problem dv/dt = 9.8 - (v/5), v(0) = 0. (a) Find the time that must elapse for the object to reach 98% of its limiting velocity. (b) How far does the object fall in the time found in part (a)?
Modify example 2 so that the falling object experiences no air resistance. (a) Write down the modified initial value problem. (b) Determine how long it takes the object to reach the ground. (c) Determine its velocity at the time of impact.
        
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Elementary and Intermediate Algebra
Elementary and Intermediate Algebra
Alan S. Tussy, R. David Gustafson 5th Edition
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This is a differential equations question. It is dropped from a height of 300m. The falling object in Example 2 satisfies the initial value problem dv/dt = 9.8 - (v/5), v(0) = 0. (a) Find the time that must elapse for the object to reach 98% of its limiting velocity. (b) How far does the object fall in the time found in part (a)? Modify example 2 so that the falling object experiences no air resistance. (a) Write down the modified initial value problem. (b) Determine how long it takes the object to reach the ground. (c) Determine its velocity at the time of impact.
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Transcript

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00:01 So here the solution for this problem is we'll get the equation as dv divided by d t.
00:07 Dv divided by d t is equals to 9 .8 multiplied with 5 minus v divided by 5.
00:17 So we'll get the equation as bv divided by 49 minus v is equals to d t divided by 5.
00:26 Where 49 minus v is equal to 0 and minus dv is equals to da so we'll get the equation as minus the a divided by a is equal to d t divided by 5 so by integrating the equation on both sides we'll get the equation as minus l n of a is equal to p divided by 5 plus so by simplifying this further we'll get the equation as ln of a is equals to minus t divided by five plus k.
01:10 So here we will get the equation as ln of 49 minus v is equals to minus t divided by minus t divided by five plus k.
01:25 But here v of 0 is equal to 0 is given so the equation will be ln of 49 minus 0 is equal to 0 plus k which means ln of 49 minus 0 is equal to k so this will come as 3 .9 is equal to k which means ln of 49 minus v is equals to minus t divided by 5 plus 3 .9.
02:04 So as for the first condition, as for the first condition, the 98 % of limiting velocity will be 49 minus v is equals to e power negative to e power negative to e.
02:29 Divided by 5 multiplied with e power 3 .9.
02:35 So this will give us the equation as 49 multiplied with e power minus t divided by 5 multiplied with 49 minus v is equal to v.
02:52 So here the limit t approaching infinity of v is equals to 49 minus 49 .4 divided by e power t divided by 5...
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