Question

Assume that a softball is thrown straight up with an initial velocity: v=vo experiences a drag force due to air resistance given by: Farag = - bv a. Use F = ma to show that the appropriate differential equation is given by: mat= -mg - kmy [where k = 1 b. Show by direct substitution that: v(t) = -9 + [kvote-kt sat isfies the differential equation. C. Assuming the ball is thrown straight up with velocity v = vo, find the time t when the ball will reach its highest point and start to fall down (leave your answer in terms of g, k, and vo). [Hint the ball will "stop at the top" i.e. v = at the highest point]. d. How high above its starting point will it go before it starts to come down again? (leave your answer in terms of g,k,and vo). e. Extra credit: Compare your answer to part d to the maximum height when b=0 (no frictional force) to find out how much energy is lost to friction on the way up to the top of the ball's trajectory when b 0.

Answer 1

Part.(a)

When a ball is projected upwards in gravitational field, forces acting on the ball are gravitational force and drag force due to air resistance

Hence net force F acting on the ball , F = - mg - bv ......................(1)

where m is mass of ball , g is acceleration due to gravity, b is drag coefficient.

Let k = b/m , then eqn.(1) is written as ,F = - m g - k m v ...............(2)

According to newton's law , net Force acting on object equals product of mass $\times$ acceleration

hence we get, ........................(1)

mx = -mg-kmu

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Part(b)

.........................(2)

-9 v(t) = 2 kv. +9.-kt + k

$\frac{dv}{dt} = -( k v_{o} + g)e^{-kt}$ ..........................(3)

Substituting for v(t) and dv/dt using eqn.(2) and eqn.(3), we check eqn.(1) as given below

$m\times \left [ -(kv_{o}+g)e^{-kt} \right ] = -mg -km\left ( \frac{-g}{k} + \frac{kv_{o}+g}{k}e^{-kt}\right )$

we check from above expression, LHS equals RHS

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Part(c)

Speed as a function of time is given as

-9 v(t) = 2 kv. +9.-kt + k

when the ball reaches maximum height its speed will be zero

Hence from above equation for velocity we get,

$0 = \frac{-g}{k} + \frac{kv_{o}+g}{k}e^{-kt}$

$e^{-kt} = \frac{g}{kv_{o}+g}$   .............................(4)

$t = \frac{1}{k}\times ln\left ( \frac{kv_{o}+g}{g} \right )$ ................(5)

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Part (d)

$v(t) = \frac{dS}{dt} = -\frac{g}{k} + \frac{kv_{o}+g}{k}e^{-kt}$

$dS = -\frac{g}{k} dt+ \frac{kv_{o}+g}{k}e^{-kt}dt$

By inegrating on both side using the limits t = 0 to t = tmax , where tmax is as given by eqn.(5)

$S = \int ds = -\frac{g}{k}\int_{0}^{t_{max}} dt + \frac{kv_{o}+g}{k}\int_{0}^{t_{max}}e^{-kt}dt$

$S = \frac{-g}{k}\times t_{max} + \frac{kv_{o}+g}{k^{2}}\left [ 1 - e^{-k t_{max}} \right ]$

if we substitute tmax from eqn.(5) and e-ktmax from eqn.(4) in the above expression for S, we get

$S = \frac{-g}{k^{2}}ln\left ( \frac{kv_{o}+g}{g} \right ) + \frac{kv_{o}+g}{k^{2}}\left [ 1 - \frac{g}{kv_{o}+g} \right ]$

after simplification, we get

$S = \frac{-g}{k^{2}}ln\left ( \frac{kv_{o}+g}{g} \right ) + \frac{v_{o}}{k}$ .......................(6)

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Part(e)

when friction due to air rsistance is absent, maximum height reachable by ball, S = vo2 / (2g)

Hence using eqn.(6)

difference in distance is given by

$\Delta S = \frac{v_{o}^{2}}{2g} - \frac{-g}{k^{2}}ln\left ( \frac{kv_{o}+g}{g} \right ) - \frac{v_{o}}{k}$

energy dissipated due to friction = equivalent potential energy of distance $\Delta$ S

energy dissipated due to friction = mg$\Delta$S