Q5-40E - University Physics with Modern Physics

Question

You throw a baseball straight upward. The drag force is proportional to $v^2$ . In terms of g, what is the y-component of the ball’s acceleration when the ball’s speed is half its terminal speed and 

 (a) it is moving up?

 (b) It is moving back down?

Step-by-Step Solution

Answer

(a) The ball’s acceleration when it is moving up is,$\frac{5}{4}g$ .

(b) The ball’s acceleration when it is moving back down is, $\frac{3}{4}g$. 

Step 1: Identification of the given data

The given data can be listed below as,

• The drag force is directly proportional to $v^2$.

Step 2: Significance of drag force

The drag force which is also known as air resistance, it is the force that act on an object opposite to its relative motion while object is moving in air.

Step 3: Determination of the ball’s acceleration when it is moving up.

Part (a)

 

The expression for drag force according to Newton’s second law when ball is moving up can be expressed as,

 $\sum F_y=f+w$

Here $f$ is drag force, is the weight.

 

Substitute $ma$ for ,$F_y$ and $\frac{1}{4}mg$ for $f$, and $mg$for $w$ in the above equation.

 $\begin{array}{c}ma=\frac{1}{4}mg+mg\\ ma=\frac{5}{4}mg\\ a=\frac{5}{4}g\end{array}$

Hence, required acceleration of ball when it is moving up is, .$\frac{5}{4}g$

Step 4: Determination of the ball’s acceleration when it is moving back down.

Part (b)

 

The expression for drag force according to Newton’s second law when ball is moving back down can be expressed as,

 $\sum F_y=w-f$

Substitute $ma$ for $F_y$, and $\frac{1}{4}mg$ for $f$, and $mg$ for $w$ in the above equation.

 $\begin{array}{c}ma=mg-\frac{1}{4}mg\\ ma=\frac{3}{4}mg\\ a=\frac{3}{4}g\end{array}$

Hence, required acceleration of ball when it is moving back down is,$\frac{3}{4}g$ .