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CHAPTER 3 - 13

# CHAPTER 3 - 13 - (c x=v t=(51.2m/s(cos44.5(7.32s)=267m 0x(d...

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( c )We find the horizontal distance from x  =  v 0 x t  = (51.2 m/s)(cos 44.5 ° )(7.32 s) =         267 m . ( d ) The horizontal velocity will be constant:  v x  =  v 0 x  = (51.2 m/s) cos 44.5 °  = 36.5 m/s. We find the vertical velocity from v y  =  v 0 y  +  a y t  = (51.2 m/s) sin 44.5 °  + (– 9.80 m/s 2 )(1.50 s) = 21.2 m/s. The magnitude of the velocity is v  = ( v x 2  +  v y 2 ) 1/2  = [(36.5 m/s) 2  + (21.2 m/s) 2 ] 1/2  =          42.2 m/s . We find the angle from tan  θ  =  v y / v x  = (21.2 m/s)/(36.5 m/s) = 0.581, which gives         θ  = 30.1 °  above the horizontal . 40. ( a ) We choose a coordinate system with the origin at the  base of the cliff, with  x  horizontal and  y  vertical, with  the positive direction up.   We find the time required for  the fall from the vertical motion: y  =  y 0  +  v 0 y t  +  !

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CHAPTER 3 - 13 - (c x=v t=(51.2m/s(cos44.5(7.32s)=267m 0x(d...

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