Solved: Vertical Motion An object is dropped from a height of 400 feet. (a) Find the

Chapter 5, Problem 96

(choose chapter or problem)

An object is dropped from a height of 400 feet.

(a) Find the velocity of the object as a function of time (neglect air resistance on the object).

(b) Use the result in part (a) to find the position function.

(c) If the air resistance is proportional to the square of the velocity, then \(d v / d t=-32+k v^{2}\), where -32 feet per second per second is the acceleration due to gravity and k is a constant. Show that the velocity v as a function of time is \(v(t)=-\sqrt{32 / k} \tanh (\sqrt{32 k} t)\) by performing \(\int d v /\left(32-k v^{2}\right)=-\int d t\) and simplifying the result.

(d) Use the result of part (c) to find \(\lim _{t \rightarrow \infty} v(t)\) and give its interpretation.

(e) Integrate the velocity function in part (c) and find the position s of the object as a function of t. Use a graphing utility to graph the position function when k = 0.01 and the position function in part (b) in the same viewing window. Estimate the additional time required for the object to reach ground level when air resistance is not neglected.

(f) Give a written description of what you believe would happen if k were increased. Then test your assertion with a particular value of k.

Text Transcription:

d v / d t=-32+k v^2

v(t)=-sqrt 32 / k

sqrt 32 k t

int d v /(32-k v^2)=-int d t

lim _t rightarrow infty v(t)