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The acceleration of a free falling body is described by a = 9.81 [1-10-4. v²] m/sec², where v is in m/sec and the positive direction is downward. If the body is released from rest at a high altitude, calculate the velocity when t = 5 sec in m/sec using the formula below. dx = 1 Ina+ x \ + C 1 (a+x) 2 S a²-x² 2 2a for a²x² 2 , where a and care constants

The acceleration of a free falling body is described by a = 9.81 [1-10-4. v²] m/sec², where v is in m/sec and the positive direction is downward. If the body is released from rest at a high altitude, calculate the velocity when t = 5 sec in m/sec using the formula below. dx = 1 Ina+ x \ + C 1 (a+x) 2 S a²-x² 2 2a for a²x² 2 , where a and care constants

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The acceleration of a free falling body is described by a = 9.81 [1-10-4. v²] m/sec², where v is in m/sec and the positive direction is downward. If the body is released from rest at a high altitude, calculate the velocity when t = 5 sec in m/sec using the formula below. I In/a+x\ + C dx 1 2 2 S a²-x² 2a la-x, for a²x², where a and c are constants .
Transcribed Image Text:The acceleration of a free falling body is described by a = 9.81 [1-10-4. v²] m/sec², where v is in m/sec and the positive direction is downward. If the body is released from rest at a high altitude, calculate the velocity when t = 5 sec in m/sec using the formula below. I In/a+x\ + C dx 1 2 2 S a²-x² 2a la-x, for a²x², where a and c are constants .
Expert Solution
Step 1

Given that,

the acceleration of the body is

a=9.811-10-4v2m/s2

 

we have to find the velocity at t=5s,

 

Since the acceleration is (v is the velocity of the body)

a=dvdt

and velocity is

v=dxdt

Then

dv=adt

steps

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