To determine The time taken by the ball to rise the remaining distance to the surface Explanation The figure shown below shows the free body diagram of the tennis ball. Figure-(1) Write the expression for the volume of the ball. V B = π D 3 6 (I) Here, the diameter of the ball is D . Write the expression for the projected area of the ball. A = π 4 D 2 (II) Write the expression for the force acting on the ball due to buoyancy. F B = ρ g V B (III) Here, the density of water is ρ the acceleration due to gravity is g , the volume of the ball is V B . Write the expression for the balance the forces action on the ball. F D = F B − W (IV) Here, the drag force acting on the ball is F D , the weight of the tennis ball is W and the buoyancy force acting on the ball is F B . Write the expression for the drag force. F D = 1 2 ρ A U 2 ( C D ) (V) Here the projected area of the ball is A the drag coefficient is C D and the terminal velocity is U . Write the expression for the Reynolds number. R e = U D v (VI) Here, the kinematic viscosity of water is v . Write the expression for the distance of the ball that needs to travel to reach the surface after it reached the terminal velocity. L = L 1 − L 2 (VII) Here, the depth of the swimming pool is L 1 , the distance when ball achieve terminal velocity is L 2 . Write the expression for the time taken by the ball. t = L U (VIII) Here the terminal velocity is U . Conclusion: Substitute 38.1 mm for D in Equation (I). V B = π ( 38.1 mm ) 3 6 = π ( 38.1 mm ( 1 m 1000 mm ) ) 3 6 = 1.734 × 10 − 4 π m 3 6 = 2.896 × 10 − 5 m 3 Substitute 38.1 mm for D in Equation (II). V B = π 4 ( 38.1 mm ) 2 = π 4 ( 38.1 mm ( 1 m 1000 mm ) ) 2 = 1.14 × 10 − 3 m 2 Refer to the table 1.6 “Approximate physical properties of some common liquids” to obtain the value of density as 999 kg / m 3 , kinematic viscosity as 1.12 × 10 − 6 m 2 / s . Substitute 2.896 × 10 − 5 m 3 for V B , 999 kg / m 3 for ρ 9.81 m / s 2 for g in Equation (III). F B = 999 kg / m 3 × 2.896 × 10 − 5 m 3 × 9.81 m / s 2 = 0.2838 kg ⋅ m / s 2 ( 1 N 1 kg ⋅ m / s 2 ) = 0.2838 N Substitute 0.2838 N for F B and 0.0245 N for W in Equation (IV). F D = 0.2838 N − 0.0245 N = 0 .2593 N Substitute 999 kg / m 3 for ρ , 1.14 × 10 − 3 m 2 for A and 0

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Author: HOCHSTEIN

Publisher: JOHN WILEY+SONS INC.

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Chapter 9.3, Problem 57P

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The time taken by the ball to rise the remaining distance to the surface

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Chapter 9.3, Problem 56P

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EBK MUNSON, YOUNG AND OKIISHI'S FUNDAME

Ch. 9.1 - Prob. 1P Ch. 9.1 - Assume that water flowing past the equilateral...Ch. 9.1 - Repeat Problem 9.1 if the object is a cone (made...Ch. 9.1 - Prob. 4P Ch. 9.1 - Prob. 5P Ch. 9.1 - Prob. 6P Ch. 9.1 - Prob. 8P Ch. 9.1 - Typical values of the Reynolds number for various...Ch. 9.1 - Prob. 11P Ch. 9.1 - Consider the following cases. (a) A small...

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