Explanation Show the free diagram of the cranberry and barrier. Write the expression of the gravitational potential energy. GPE= V 1 = m g h (I) Here, mass of the crate is m , acceleration due to gravity is g and the height above the ground is h . Write the expression of the conservation of energy. T 1 + V 1 = T 2 + V 2 T 1 + V 1 = 1 2 m C ( v C ) 1 2 + V 2 (II) Determine the conservation of momentum of cranberry with respect x’ -axis direction. m C ( v C x ′ ) 1 = m C ( v C x ′ ) 2 (III) Write the formula for coefficient of restitution ( e ) along y ′ − a x i s . e = ( v P ) 2 − ( v C y ′ ) 2 ( v C y ′ ) 1 − ( v P ) 1 (IV) Here, the velocity of the dinner plate is ( v P ) , the velocity of the cranberry is ( v C y ′ ) Write the formula for velocity of the cranberry along y − a x i s . v y = ( v 0 ) y + a C t (V) Here, the final velocity of the cranberry along y − a x i s is v y , the initial velocity of the cranberry along y − a x i s is ( v 0 ) y , the acceleration due to gravity of cranberry is a C , and the time is t . Write the formula for position of the cranberry along y − a x i s . s y = ( s 0 ) y + ( v 0 ) y t + 1 2 ( a C ) y t 2 (VI) Here, the final position of the cranberry along y − a x i s is s y , the initial position of the cranberry along y − a x i s is ( s 0 ) y , the initial velocity of the cranberry along y − a x i s is ( v o ) y , the acceleration due to gravity is ( a C ) y , and the time is t . Write the formula for velocity of the cranberry along x − a x i s . s x = ( s 0 ) x + v x t (VII) Here, the horizontal velocity of the cranberry along x − a x i s is v x , the initial velocity of the cranberry along x − a x i s is ( s 0 ) x , the acceleration due to gravity of cranberry is v x , and the time is t . Determine the height of the barrier. h = s y + sin θ d (VIII) Conclusion: Refer Figure (1). The weight of cranberry is W = m g Substitute 1.75 m for h , W for m g in Equation (I). V 1 = W ( 3.5 ) = 3.5 W Substitute 0 for T 1 , 3.5 W for V 1 , and ( W 32.2 ) for m C in Equation (III). 0 + 3.5 W = 1 2 ( W 32.2 ) ( v C ) 1 2 + 0 ( v C ) 1 = 15.01 ft/s (VIII) Substitute ( 15.01 ) sin θ for ( v C x ′ ) 1 , and ( v C ) 2 cos ϕ for ( v C x ′ ) 2 in Equation (IV) m C ( 15

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ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Chapter 15, Problem 90P

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The dimension of the barrier, d .

The height of the barrier, h .

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Chapter 15, Problem 89P

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The dimension of the barrier, d . The height of the barrier, h .

Solution Summary: The author shows the free diagram of the cranberry and barrier. Write the expression of gravitational potential energy.