An object falling from rest at a height of y = 4000m accelerates at g 9.8 m/s2.

Question

An object falling from rest at a height of y = 4000m accelerates at g 9.8 m/s2. Its downward motion is opposed by a friction force that we will assume is proportional to the square of the velocity with constant . Identify the terms in the equation of motion (including the initial conditions) of the object What are the units of ? Terminal velocity occurs when the object stops accelerating (as friction force balances gravity). Compute the terminal velocity in terms of g and c. If the terminal velocity Is y'(t) = 200km/h, compute (Ensure you are using the correct units). Use a numerical method to solve for the elevation and speed as functions of time. How close will it get to the terminal velocity before it reaches the ground? The equation for u = dy/dt can be solved exactly as it is a first-order DE. Solve for the velocity and verify your conclusions in (c).

Explanation / Answer

I am sure you will agree that this is four different questions. I'll work on the first two. (a) y'' represents acceleration of the object. -g represents the acceleration of gravity. epsilon (y')^2 gives air resistance proportional to the square of the velocity. The initial conditions y(0) = 4000, y'(0) = 0 tell you that the object is dropped (zero initial velocity) from a height of 4000 meters. Note that the units of epsilon (y')^2 must give an acceleration. Since (y')^2 gives m^2/s^2 (velocity squared) the units of epsilon must be m^{-1}, or per meter. (b) let y' = u. Then terminal velocity is when -g + epsilon u^2 = 0, or when u = sqrt(g/epsilon). If the terminal velocity is given at 200 km/hr = 55.6 m/s, then the value of epsilon satisfies epsilon = g/u^2 = 9.8 (m/s^2) / 55.6^2 (m^2/s^2) = 0.0032 per meter Good luck on the rest.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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