Find the maximum height?

Question

A stone weighing 1/2 lb is thrown upward from an initial height of 5 ft with an initial speed of 32 ft/s. Air resistance is proportional to speed, with k=1/128 lb-s/ft. Find the maximum height attained by the stone.

 

Answer: 17.10 ft

 

Please show all your work step by step. This is a differential equations problem.

Answer 1

To find the differential equation we need to set up Newton's Second Law

-W -k v = m a

Where W is the weight and the negative sign in kv comes from drag force acting in the opposite direction as v.

-W/m = dv/dt + k/m v

This is a differential equation that can be solved by integrating factor.

-W/m e^{k/m t} = d/dt( e^{k/m t} v)

-W/k + c e^{-k/m t} = v(t)

Applying the initial conditions

v(t) = -W/k + (W/k + v₀) e^{-k/m t}

The time at its peak should be

t = -m/k ln(W/(W+kv₀)).

Notice that the mass m should be W/g = 1/2 lbs / 32ft/s²

v(t) = -64 ft/s + 96 ft/s e^{- t/2s}

v(t_max) = 0 → t_max = -2 ln(64/96)

Δx = x_f - x_i = x_f - 5 ft = ∫₀^-2ln(64/96) -64 + 96 e^-t/2 dt = 12.10

x_f = 17.10