A bullet of mass m, fired straight up with an initial velocityof v0, is slowed by the

Chapter 8, Problem 62

(choose chapter or problem)

A bullet of mass m, fired straight up with an initial velocity of \(v_{0}\), is slowed by the force of gravity and a drag force of air resistance \(k v^{2}\), where k is a positive constant. As the bullet moves upward, its velocity v satisfies the equation

\(m \frac{\partial y}{d t}=-\left(k v^{2}+m g\right)\)

where g is the constant acceleration due to gravity.

(a) Show that if \(x=x(t)\) is the height of the bullet above the barrel opening at time t, then

\(m v \frac{d y}{d x}=-\left(k v^{2}+m g\right)\)

(b) Express x in terms of v given that \(x=0\) when \(v=v_{0}\)

(c) Assuming that

\(v_{0}=988 \mathrm{~m} / \mathrm{s}, \mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2} \mathrm{~m}=3.56 \times 10^{-3} \mathrm{~kg}, k=7.3 \times 10^{-6} \mathrm{~kg} / \mathrm{m}\)

use the result in part (b) to find out how high the bullet rises. [Hint: Find the velocity of the bullet at its highest point.]

Equation Transcription:

 

Text Transcription:

v_0

kv^2

m dv/dt=-kv^2+mg

x=x(t)

mv dv/dx=-kv^2+mg

x=0

v=v_0

 v_0=988 m/s, g=9.8 m/s^2  m=3.56x10^-3 kg, k=7.3 x 10^-6kg/m