= _____________m/s Solution vi=1.5m/sec 5.52g-T=5.52a-----------2 T=50.3a-------

Question

= _____________m/s

Explanation / Answer

vi=1.5m/sec 5.52g-T=5.52a-----------2 T=50.3a------------1 1+2 5.52g=(55.82)a =>a=5.52*9.8/55.82 =>a=.969m/s2 h=14.9m Vy^2-uy^2=2as Vy^2-1.5^2=2*.969*14.9 Vy=2.25+2*.969*14.9 =>Vy=5.58m/sec velocity at the time before time hitting ground Vy-uy=at 5.58-1.5=.969t =>t=4.21sec assume the man reaches the edge of cliff when Vy of block is 5.58m/sec uman=1.5m/sec a=.969m/s2 so Vman=uman+at =1.5+4.21*.969 =5.58m/sec so Vmanx=5.58m/sec Vmany inital =0 Vman final ^2-uman y initial ^2 =2gh h=26.8m Vman y final^2=0+2*9.8*26.8 Vman y final=22.92m/sec velocity of man at the bottom of cliff is 5.58i-22.92j magnitude of man's velocity =23.59m/sec

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