differential equation the velocity u (in meters per secold, suppse the ngeutont

Question

differential equation

the velocity u (in meters per secold, suppse the ngeutont (a) Find the velocity and distance fallen at time t seconds. (b) Find the time at which the velocity is one-fifth of the limiting velocity (29) A bullet weighing lou is fired vertically downward from a stationary helicopter with a muale velocity of 1200n/sec The air resistance (in pounds) is numerically equal to 16-2, where o is the velocity (in feet per second). Find the velocity of the bullet as a function of time. (30) A case of canned milk weighing 24lb is released from rest at the top of a plane metal slide which is 30t long and inclined 45° to the horizontal. The air resistance (in pounds) is numerically equal to one-third the velocity (in feet per second). The coefficient of friction is 0.4 (a) Find the velocity of the moving case 1 second after it is released. (b) What is the velocity when the case reaches the bottom of the slide?

Explanation / Answer

given initial velociy of bullet U = 1500ft/sec air resistance Fa =16^(-5)( V^2) velocity of bullet at any time

equation of motion in free fall under air resistance is given by, F = mg-Fa

force acting on the body is given by , F=m(dv/dt)

   m(dv/dt) = mg - (16^-5)V^2

dv/dt = g - (16^-5)V^2/m

dv/dt = g (1-((16^-5)V^2)mg)

let (mg/16^-5) = c

dv/dt = g(1-V^2/c)

dv/dt = (g/c) ( c-V^2)

dv/(c-V^2) = g/c

   taking (1-(v^2/c)) =x

( (c^0.5)/2 ) dx/x.sqrt(1-x) =-g/c

   integrating on both sides

   arc tan x = g/c^(3/2)t

   arctan (1-V^2/c) =(g/c^1.5)t

V = SQRT( c(1-tan(gt/c^1.5))) where c=mg/16^-5

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.