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 d2y/dt2 = - 9 + (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

d^2y/dt^2 is acceleration g is acceleration due to gravity dy/dt is velocity unit of epsilon is metre^-1

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