A baseball has a terminal speed of 42 m/s in air (ρ = 1.2 kg/m³).What would be its terminal speed in water (ρ = 1.0 x 10³ kg/m³)? A)0.05 m/s B)1.5 m/s C)18 m/s D)42 m/s E)1200 m/s

Question

Quizplus · Accepted Answer

Terminal velocity is given by the formula $v_t = \frac{2mg}{ho A C_d}$, where $m$ is the mass of the object, $g$ is the acceleration due to gravity, $ho$ is the density of the fluid, $A$ is the cross-sectional area of the object, and $C_d$ is the drag coefficient. Since the baseball has the same size and shape in both air and water, we can assume that $A$ and $C_d$ are constant. We can therefore write:

$\frac{v_{t,water}}{v_{t,air}} = \frac{ho_{air}}{ho_{water}}$

Substituting the given values, we have:

$\frac{v_{t,water}}{42} = \frac{1.2}{1.0 	imes 10^3}$

Solving for $v_{t,water}$, we get:

$v_{t,water} = 1.5$ m/s.

Therefore, the answer is B.

A Baseball Has a Terminal Speed of 42 M/s in Air