Explanation Given: The velocity υ A is 80 ft/s . The distance s x is 35 ft . The distance s y is − 20 ft . The velocity υ A and distance is shown in Figure (I). Write the expression for final distance in x direction. s x = ( s 0 ) x + ( υ A ) x t (I). Here, final distance in the horizontal direction is s x , initial distance in the horizontal direction is ( s 0 ) x , initial velocity in the horizontal direction is ( υ A ) x and time is t . Write the expression for final distance in y direction. s y = ( s 0 ) y + ( υ A ) y t + 1 2 a c t 2 (II). Here, final distance in the vertical direction is s y , initial distance in the vertical direction is ( s 0 ) y , initial velocity in the vertical direction is ( υ A ) x , acceleration due to gravity is a c and time is t . From Figure (I) write the expression for ( υ A ) x . ( υ A ) x = υ A cos θ (III). From Figure (I) write the expression for ( υ A ) y . ( υ A ) y = υ A sin θ (IV). Conclusion: Substitute 80 ft/s for υ A in Equation (I). ( υ A ) x = υ A cos θ = 80 cos θ Substitute 80 ft/s for υ A in Equation (I). ( υ A ) y = υ A sin θ = 80 sin θ From Figure (I) substitute 35 ft for s x , 0 for ( s 0 ) x and 80 cos θ for ( υ A ) x in Equation (I). s x = ( s 0 ) x + ( υ A ) x t 35 = 0 + 80 cos θ t t = 0.43 cos θ (V) From Figure (I) substitute 0.43 cos θ for t , − 80 sin θ for ( υ A ) y , − 20 ft for s y , 0 for ( s 0 ) y and − 32

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Chapter 12, Problem 107P

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The two possible angles θ1 and θ2 .

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Chapter 12, Problem 106P

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The two possible angles θ 1 and θ 2 .

Solution Summary: The author illustrates the possible angles of theta and the distance in Figure (I).

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