A heavy ball is retarded by air resistance which is modelled by kv², where v ms^-1 is the speed of the ball and k is a positive constant. When falling under gravity in air, its terminal velocity is V ms^-1. i) Write an expression for the total force on the ball when it is

Question

A heavy ball is retarded by air resistance which is modelled by kv², where v [latex]ms^{-1}[/latex] is the speed of the ball and k is a positive constant. When falling under gravity in air, its terminal velocity is V [latex]ms^{-1}[/latex].
i) Write an expression for the total force on the ball when it is descending and hence derive an expression for V² in terms of k, g and the mass of the ball.
ii) The ball is projected vertically upwards with an initial speed u [latex]ms^{-1}[/latex].
Derive a differential equation, involving s and V, but not k, covering its ascent.
iii) Show that, when u = 50 and V = 100, the ball reaches a height of almost 114 m.

Accepted Answer

i) Assume that the mass of the ball is m and the resistance is kv². When descending, the net downward force is mg - kv².

The terminal velocity, V, is obtained when the force is zero, so mg = kV².

⇒ [latex]V^{2} = \frac{mg}{k}[/latex]

ii) When ascending, both the weight and the air resistance are in the same direction.
Take the positive direction as upwards, so the forces are negative and the initial velocity is positive.

The equation of motion is then

[latex]\begin{matrix}- mg - kv^{2} = mv \frac{dv}{ds} & \boxed{ ext{For an equation in s write }a = v\frac{dv}{ds}.} \end{matrix} [/latex]

[latex] - \frac{mg}{k } - v^{2} = \frac{m}{k} v \frac{dv}{dx}.[/latex]

Substituting [latex]\frac{mg}{k} = V^{2}[/latex] gives

[latex]- V^{2} - v^{2} = \left(\frac{V^{2}}{g} ight)v \frac{dv}{ds}[/latex]

[latex]v\frac{dv}{ds} = \frac{-g(V^{2} + v^{2})}{V^{2}}[/latex].

iii) Separating the variables and integrating gives

[latex]\frac{v dv}{V^{2} + v^{2}} = - \frac{g}{V^{2}} ds[/latex]

[latex]\frac{1}{2} ln(V^{2} + v^{2}) = \frac{gs}{V^{2}} + c.[/latex]

If v = u when s= 0, [latex]c = \frac{1}{2}(V^{2} + u^{2})[/latex]

⇒ [latex]\frac{gs}{V^{2}} = \frac{1}{2} ln(V^{2} + u^{2}) - \frac{1}{2} ln(V^{2} + v^{2})[/latex]

[latex]\begin{matrix} s = \frac{V^{2}}{2g} ln\left(\frac{V^{2} + u^{2}}{V^{2} + v^{2}} ight). & \boxed{ ext{This is the general solution for s in terms of v.}} & \boxed{ \begin{matrix} ext{It is useful to check the dimensions. The expression} \ ln\left(\frac{V^{2} + u^{2}}{V^{2} } ight) ext{ is dimensionless so, on the right-hand side, }\ \frac{V^{2}}{2g} ext{ gives } \frac{(LT^{-1})^{2}}{LT^{-2}} = \ L^{2}T^{-2}L^{-1}T^{2} = L, ext{as required.}\end{matrix}} \end{matrix}[/latex]

When v = 0, s = h so

[latex]h = \frac{V^{2}}{2g}ln\left(\frac{V^{2} + u^{2}}{V^{2} } ight).[/latex]

Substituting the values V = 100, u = 50 given in the question:

[latex]h = \frac{1002}{2g} ln 1.25 = 113.8.[/latex]

The ball reaches a height of 113.8 m.

Question 18.9: A heavy ball is retarded by air resistance which is modelled...

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