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Baseball is dropped from empire state building.

Baseball radius = 0.0366m

Baseball mass = 0.145kg

Empire state building height = 381m

Calculate the initial potential energy and final kinetic energy.

I know the potential energy is U = mgh = (mass)(height)(gravity)

I also know kinetic energy is KE = (1/2)(m)(v^2)

I do not know however how to take air resistance into account when solving for KE. So basically, I do not know how to get the velocity.

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The ball experiences a drag force due its movement through the air. Under certain conditions (that we'll assume) this drag force is adequately modeled by the so called drag equation $$ F_D=\frac{1}{2}\rho C_DAv^2 $$ where

  • $F_D$ is the drag force, which is by definition the force component in the direction of the flow velocity
  • $\rho$ is the mass density of the fluid (air in our case: $\rho\approx 1.1839$ Kg/m$^3$ at $1$ atm and 25 °C)
  • $v$ is the velocity of the ball relative to the air
  • $A$ is the reference area, which in our case is just the cross sectional area of the ball: $A=\pi r^2$
  • $C_D$ is the drag coefficient - a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag; in the case of a sphere, $C_D=0.47$

Under this model, the ball eventually reaches a terminal speed that we can find from Newton's second law (working on a single axis, say $y$): $$ \sum F_y=-mg+Dv^2=ma_y $$ where for convenience I have defined $D=1/2\rho C_DA$. Terminal speed ($v_t$) is defined as the speed of the ball for which $a_y=0$; therefore $$ \begin{align} -mg+Dv^2_t &= 0 \\ v_t &= \sqrt{\frac{mg}{D}} \end{align} $$ which result to be $v_t\approx 34.837$ m/s. With this we can answer your question:

  • $U_\text{initial}=mgh\approx541.401$ J
  • $K_\text{final}=\frac{1}{2}mv_t^2=\frac{m^2g}{2D}\approx87.991$ J

However, there is a problem, because we should not assume a priori that the ball hits the ground near this terminal speed. Instead, we will find the exact moment when it reaches the floor, and then see what the value of the speed at that instant. Let's work with the Newton's second law again $$ \begin{align} -mg+Dv^2_t &= m\frac{dv}{dt} \\ \frac{D}{m}\int_0^tdt' &= \int_0^v\frac{dv'}{v'^2-v_t^2} \\ \frac{D}{m}t &= -\frac{1}{v_t}\tanh^{-1}\left(\frac{v}{v_t}\right) \end{align} $$ from where $$ v(t) = -v_t\tanh\left(\frac{Dv_t}{m}t\right) $$ The minus sign is because we are working with $v$ as component; Just in case you're wondering, here is the graph of the previous expression:

enter image description here

For the position as function of time, $$ \begin{align} \int_0^tv(t')dt' &= \int_h^ydy' \\ -v_t\int_0^tdt'\tanh\left(\frac{Dv_t}{m}t\right) &= y-h \\ -v_t\cdot\frac{m}{v_tD}\ln\left(\cosh\left(\frac{Dv_t}{m}t\right)\right) &= y-h \end{align} $$ from where $$ y(t) = h-\frac{m}{D}\ln\left(\cosh\left(\frac{Dv_t}{m}t\right)\right) $$ Also, here is the graph of $y(t)$:

enter image description here

Now, we want to know the instant $t_\text{final}$ such that $y(t_\text{final})=0$, so \begin{align} t_\text{final} &= \frac{m}{Dv_t}\cosh^{-1}\left(\exp\left(\frac{hD}{m}\right)\right) \\ &\approx 13.3914\; \text{s} \end{align} Therefore, the speed of the ball when it hits the ground will be $$ \begin{align} v(t_\text{final}) &= v_t\tanh\left(\frac{Dv_t}{m}t_\text{final}\right) \\ &\approx 34.7996\; \text{m/s} \end{align} $$ which it turns out to be quite close to the value $v_t$. Finally, Of course the value $U_\text{initial}$ it doesn't changes, while the value of $K_\text{final}$ changes very little: $$ K_\text{final} \approx 87.7954\; \text{J} $$

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  • $\begingroup$ In accordance with our homework policy, I'm temporarily deleting this. $\endgroup$ – Qmechanic Oct 3 '16 at 14:35
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You can find this in many books; for example:

enter image description here

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  • $\begingroup$ If you have troubles with the drag equation, the first to do is to solve the simplest case, for example doing the retarding force proportional to the velocity which means you will have a term k*dy/dt, besides your mg in the y direction to start. $\endgroup$ – Cristian Allendes Flores Oct 5 '16 at 18:22

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