Respuesta :
Answer:
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:
[tex]\frac{dv}{(0.00025*v^2 + 32)} = -dt\\\\\frac{1}{32}*\frac{dv}{(\frac{v^2}{128000} + 1)} = -dt.[/tex]
- Recall the basic integration:
[tex]tan^(^-^1^) (x) = \int\limits^a_b {\frac{1}{(1 + x^2)}} \, dx[/tex]
- So we can write:
[tex]\frac{1}{32}*tan^(^-^1^) (\frac{v}{357.77087} ) = \frac{1}{32} \int\limits^a_b {\frac{1}{(1 + \frac{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 )[/tex]
- Where A is the integration constant to be evaluated by the initial value given v(0) = 620 ft/s. We compute:
[tex]A = tan^-^1 ( 0.00279508*620 ) = 1.04742\\[/tex]
- The solution to the ODE becomes:
[tex]v (t) = 357.77087*tan( -0.08944t + 1.04742)[/tex]
- 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):
[tex]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[/tex]
