To determine
a)To calculate: The limit limt→∞v, where v=mgc(1−e−ct/m)
Answer
limt→∞v=mgc
Explanation
We compute the limit as t→∞
limt→∞v=limt→∞mgc(1−e−ct/m)=mgclimt→∞(1−e−ct/m)=mgc(1−e−∞)=mgc(1−0)=mgc
We Know that as x→∞⇒e−x→0
Final statement: We can conclude that as time increases velocity of the object depends on its mass, acceleration due to gravity and gravitational constant
limt→∞v=mgc
To determine
b)To calculate:For fixed t, calculate limc→0+v
Explanation
limc→0+v=limc→0+mgc(1−e−ct/m)=mglimc→0+1c(1−e−ct/m) (00 Form)Applying l'Hospital's Rule, we get=mglimc→0+−e−ct/m(−t/m)1=mg(t/m)=gt
We can conclude that in a vacuum a falling object would continue to accelerate and its velocity will increase indefinitely
Final statement: v=gt