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 c. Identify the terms in the equation of motion (including the initial conditions) of the object d2y / dt2 = -g + (dy / dt)2, y(0) = 4000, y'(0) = 0 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 . 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

the units of e are: units of e*(dy/dt)2 are m/s^2 so units of e ==> (m/s^2) / (m^2 / s^2) ==>m/m^2 =>1/m =>m-1 units of e ==> m-1

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