Question

When the effect of aerodynamic drag is included, the y‐acceleration of a baseball moving vertically upward is au = −g − kv2, while the acceleration when the ball is moving downward is ad = −g + kv2, where k is a positive constant and v is the speed in meters per second. If the ball is thrown upward at 30 m/s from essentially ground level, compute its maximum height h and its speed vf. upon impact with the ground. Take k to be 0.006 m−1 and assume that g is constant.

Answer 1

Answer:

Explanation:

The detailed steps and appropriate integration and differentiation is as shown in the attached files.

au = −g − kv2

ad = −g + kv2

Answer 2

The maximum height and speed respectively if the ball is thrown upward at 30 m/s are;

Maximum height = 36.55 m

Speed at maximum height; v_f = 24.09 m/s

We are given;

upward acceleration; a_u = -g - kv²

downward acceleration; a_d = -g + kv²

k = 0.006 /m

Initial speed; u = 30 m/s

Now, acceleration is;

a = v(dv/dy)

Thus;

dy = (v/a)dv

Integrating to get;

$\int\limits^h_0 {dy} \, = \int\limits^0_u \frac{u}{-g - kv^{2} } } \, du$  

Integrating with the given limits, we now have;

h = $\frac{-1}{2k}$(In (g + k(0)) - (In(g + k(30²)))

h = -¹/₍₂ ₓ ₀.₀₀₆₎(In 9.81 - (In(9.81 + (0.006 * 900)

h = -83.33(2.2834 - 2.722)

h = 36.55 m

B ) For the downward motion, we have;

Integrating to get;

$\int\limits^0_h {dy} \, = \int\limits^v_0 \frac{v}{-g + kv^{2} } } \, dv$

Integrating with the given limits, we now have;

-h = $\frac{1}{2k}$(In(-g + k(v²)))  

Thus;

-36.55 = ¹/₍₂ ₓ ₀.₀₀₆₎(In(-9.81 + 0.006v²)) - In(-9.81)

(In(-9.81 + 0.006v²)/-9.81) = -0.4386

(-9.81 + 0.006v²)/-9.81 = e⁻⁰'⁴³⁸⁶

0.645 * -9.81 = -9.81 + 0.006v²

-6.3268 =  -9.81 + 0.006v²

0.006v² = -9.81 + 6.3268

0.006v² = 3.4832

v = √(3.4832/0.006)

v = 24.09 m/s

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