Question

A 32 lb projectile is shot vertically upward with initial velocity at 160 15 ft/sec. If the air resistance is proportional to the square of the velocity, with proportionality constant .00025, then

a. Show the differential equation governing upward motion is:
dv/dt= -32- 0.00025v^2
b. Show the maximum height achieved at around t= 11.71 seconds

Answer 1

Explanation:

Given:-

- The initial speed of projectile fired, vi = 620 ft/s ... ( Mistake in question )

- The mass of the projectile, m = 32 lb

- Drag proportionality constant, C = 0.00025

Find:-

a. Show the differential equation governing upward motion is:

dv/dt= -32- 0.00025v^2

b. Show the maximum height achieved at around t= 11.71 seconds

Solution:-

- We will consider the projectile as our system and mark the forces acting on it. There two forces i.e ( Force due to gravitational pull of the earth and Drag resistive force).

- Both the drag force (D) and gravitational force (W) acts downward.

- Using Newton's 2nd Law of motion we will develop an expression as follows:

F_net = m*a

-D - W = m*(dv/dt)

- The expressions for drag and gravitational forces are:

D = C*v^2

W = m*g

Where, g = 32 ft/s^2 ..... (gravitational constant)

- Substitute the respective expressions in the Newton's second Law.

-C*v^2 - m*g = m*(dv/dt)

(dv/dt) = -(C/m)*v^2 - g

(dv/dt) = -(0.00025/(32/32))*v^2 - 32

(dv/dt) = -0.00025*v^2 - 32

- To determine the relationship for velocity (v) as a function of time (t) we will solve the above ODE.

- Separate variables:

dv / (0.00025*v^2 + 32) = -dt

- Make a trigonometric substitution visualizing ( 1 + v^2 ) term in the denominator for Left hand side. So pull out constant (32) from denominator:

`(dv)/((0.00025*v^2 + 32)) = -dt\\\\(1)/(32)*(dv)/(((v^2)/(128000) + 1)) = -dt.`

- Recall the basic integration:

`tan^(^-^1^) (x) = \int\limits^a_b {(1)/((1 + x^2))} \, dx`

- So we can write:

`(1)/(32)*tan^(^-^1^) ((v)/(357.77087) ) = (1)/(32) \int\limits^a_b {(1)/((1 + (v^2)/(128000) ))} \, dv\\\\11.1803*tan^(^-^1^) (0.00279508*v) = -t + A\\\\tan^(^-^1^) (0.00279508*v) = -0.08944*t + A\\\\0.00279508*v = tan ( -0.08944*t + A )\\\\v = 357.77087* tan ( -0.08944*t + A )`

- Where A is the integration constant to be evaluated by the initial value given v(0) = 620 ft/s. We compute:

`A = tan^-^1 ( 0.00279508*620 ) = 1.04742\\`

- The solution to the ODE becomes:

`v (t) = 357.77087*tan( -0.08944t + 1.04742)`

- The maximum height achieved by the projectile can be determined by recalling that velocity of projectile must be 0. So set v(t) = 0, and solve for time (t):

`tan^(^-^1^) (0.00279508*v) = -0.08944t + 1.047\\\\tan^(^-^1^) (0) = -0.08944t + 1.047\\\\0 = -0.08944*t + 1.047\\\\t = 1.047 / 0.08944\\\\t = 11.706 s`