Explanation Given: The mass of the ball is m = 0.5 kg . The initial velocity of the ball at A is v A = 10 m/s . The coefficient of the restitution between the ball and the wall is e = 0.5 . Draw the free body diagram of projectile motion of the ball as shown in Figure (1). Write the formula for position of the ball along x − a x i s . s x = ( s 0 ) x + ( v 0 ) x t (I) Here, s x is the final position of the ball along x − a x i s , ( s 0 ) x is the initial position of the ball along x − a x i s , ( v o ) x is the initial velocity of the ball along x − a x i s and t is the time. Write the formula for position of the ball along y − a x i s . s y = ( s 0 ) y + ( v 0 ) y t + 1 2 g t 2 (II) Here, s y is the final position of the ball along y − a x i s , ( s 0 ) y is the initial position of the ball along y − a x i s , ( v o ) y is the initial velocity of the ball along y − a x i s , g is the acceleration due to gravity and t is the time. Write the formula for velocity of the ball along y − a x i s . v y = ( v 0 ) y + g t (III) Here, v y is the final velocity of the ball along y − a x i s , ( v 0 ) y is the initial velocity of the ball along y − a x i s , g is the acceleration due to gravity and t is the time Refer Figure (1). Here, the wall does not move during the impact. Hence the velocity along y − a x i s is conserved. Write the formula for coefficient of restitution ( e ) . e = ( v B ) x 2 − 0 0 − ( v B ) x 1 (IV) Here, v B is the velocity of the ball at B , the suffix x indicates the velocity is along x − a x i s and 1 , 2 for indicates the before and after impact respectively. Write the formula for direction ( θ ) of the ball. From Figure (1). tan θ = ( v ) y ( v ) x (V) Conclusion: a) Refer Figure (1). The initial velocity of the ball along x − a x i s is ( v 0 ) x = v A cos 30 ° = ( 10 m/s ) cos 30 ° = 8.66 m/s The initial velocity of the ball along y − a x i s is ( v 0 ) y = v A sin 30 ° = ( 10 m/s ) sin 30 ° = 5 m/s The final position of the ball along x − a x i s is s x = 3 m . Calculate the time required to strike the wall at B . Substitute 3 m for s x , 0 for ( s 0 ) x and 8.66 m/s for ( v 0 ) x in Equation (I). 3 m = 0 + ( 8.66 m/s ) t t = 3 8.66 t = 0.3464 s The ball requires the time of t = 0.3464 s to strike the wall at B . Calculate the velocity of the ball along y − a x i s at the time t = 0.3464 s . Substitute ( v B ) y 1 for v y , 5 m/s for ( v 0 ) y , − 9.81 m/s 2 for g and 0.3464 s for t in Equation (III). ( v B ) y 1 = 5 m/s + ( − 9.81 m/s 2 ) ( 0.3464 s ) = 1.6018 m/s The velocity of the ball along x − a x i s at the time t = 0.3464 s is ( v B ) x 1 = ( v 0 ) x = 8.66 m/s Calculate the net velocity of the ball before striking the wall at B . ( v B ) 1 = ( v B ) x 1 2 + ( v B ) y 1 2 Substitute 8.66 m/s for ( v B ) x 1 and 1.6018 m/s for ( v B ) y 1 . ( v B ) 1 = ( 8.66 m/s ) 2 + ( 1.6018 m/s ) 2 = 77.5614 = 8.8069 m/s Thus, the velocity of the ball when it strikes the wall at B is 8.81 m/s . Refer Equation (V). Calculate the direction of the ball while strikes the wall at B . tan θ 1 = ( v B ) 1 y ( v B ) 1 x Substitute 8.66 m/s for ( v B ) x 1 and 1.6018 m/s for ( v B ) y 1 . tan θ 1 = 1.6018 m/s 8

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 15.4, Problem 81P

To determine

a) The initial velocity of the ball when it strikes the wall at B .

b) The velocity of the ball when it rebounds from the wall if e=0.5 .

c) The distance (s)

from the wall to where it strikes the ground at C .

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Chapter 15.4, Problem 80P

Chapter 15.4, Problem 82P

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Chapter 15 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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a) The initial velocity of the ball when it strikes the wall at B . b) The velocity of the ball when it rebounds from the wall if e = 0.5 . c) The distance ( s ) from the wall to where it strikes the ground at C .

Solution Summary: The author calculates the initial velocity of the ball when it strikes the wall at B, the direction of it, and the distance from it to the ground.

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a) The initial velocity of the ball when it strikes the wall at B .

b) The velocity of the ball when it rebounds from the wall if e=0.5 .

c) The distance (s)

from the wall to where it strikes the ground at C .

Want to see the full answer?

Chapter 15 Solutions

a) The initial velocity of the ball when it strikes the wall at B . b) The velocity of the ball when it rebounds from the wall if e = 0.5 . c) The distance ( s ) from the wall to where it strikes the ground at C .

Solution Summary: The author calculates the initial velocity of the ball when it strikes the wall at B, the direction of it, and the distance from it to the ground.

Concept explainers

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a) The initial velocity of the ball when it strikes the wall at B . b) The velocity of the ball when it rebounds from the wall if e=0.5 . c) The distance (s) from the wall to where it strikes the ground at C .

Want to see the full answer?

Chapter 15 Solutions

a) The initial velocity of the ball when it strikes the wall at B .

b) The velocity of the ball when it rebounds from the wall if e=0.5 .

c) The distance (s)

from the wall to where it strikes the ground at C .