problem03_61

University Physics with Modern Physics with Mastering Physics (11th Edition)

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3.61: a) The expression for the range, as derived in Example 3.10, involves the sine of twice the launch angle, and , 2 sin 2 sin 180 cos 2 cos 180 sin ) 2 180 ( sin )) 90 ( 2 ( sin 0 0 0 0 0 α = ° - ° = - ° = - ° α and so the range is the same. As an alternative, using cos ) 90 sin( 0 = - ° and 0 0 sin ) 90 cos( = - ° in the expression for the range that involves the product of the sine and cosine of
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Unformatted text preview: gives the same result. b) The range equation is g v R 2 sin 2 = . In this case, m/s 2 . 2 = v and m 25 . = R . Hence ), m/s .2 2 m)/( 25 . )( m/s 8 . 9 ( 2 sin 2 2 = or 5062 . 2 sin = ; and = 2 . 15 or 8 . 74 ....
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