ch09-p041 - i j) 41. With v0 = (9.5 + 4.0 m/s, the initial...

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Using Equation 4-26, the range of the athlete without using halteres is 2 2 00 0 2 sin 2 (10.31m/s) sin 2(22.8 ) 7.75 m. 9.8 m/s v R g θ ° == = On the other hand, if two halteres of mass m = 5.50 kg were thrown at the maximum height, then, by momentum conservation, the subsequent speed of the athlete would be 2 (2 ) x xx x Mm M mv Mv v v M + ′′ += ¡ = Thus, the change in the x -component of the velocity is 0 0 2 2 2(5.5 kg) (9.5 m/s) 1.34 m/s. 78 kg xxx x x x m vvv v v v MM + Δ = −= = = The maximum height is attained when 0 0 yy vv g t =−
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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