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

Step-by-step 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:

- Recall the basic integration:

- So we can write:

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

- The solution to the ODE becomes:

- 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):

y=mx+b

step-by-step explanation:

m slope

b y-intercept

ok

step-by-step explanation: