Question

The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial speed of 40.2 m/s (90 mph). Drag coefficient for a pitched baseball equals 0.35.
Part A: What is the magnitude of the ball's acceleration due to the drag force?


Part B: If the ball had this same acceleration during its entire 18.4 m trajectory, what would its final speed be?

Answer 1

Answer:

Part A)

Acceleration of the ball is 10.1 m/s/s

Part B)

the final speed of the ball is given as

$v_f = 35.3 m/s$

Explanation:

Part a)

As we know that drag force is given as

$F = \frac{C_d \rho A v^2}{2}$

$C_d = 0.35$

$A = \frac{\pi d^2}{4}$

$A = \frac{\pi(0.074)^2}{4}$

$A = 4.3 \times 10^{-3} m^2$

$v = 40.2 m/s$

so we have

$F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}$

$F = 1.46 N$

So acceleration of the ball is

$a = \frac{F}{m}$

$a = \frac{1.46}{0.145}$

$a = 10.1 m/s^2$

Part B)

As per kinematics we know that

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 40.2^2 = 2(-10.1)(18.4)$

$v_f = 35.3 m/s$