A body is left free at height h= 19.6m ( with initial velocity v0= 0 ) . What is the path covered by the body . 1) in initial time interval t1 = 0.1 s
2) and during the time period t2 = 0.1 s just before striking the ground ?
1
Expert's answer
2018-03-16T08:36:07-0400
1) The path covered by the body is S_1=(gt_1^2)/2=(9.8⋅〖0.1〗^2)/2=0.049m. 2) Time of falling is t_0=√(2h/g)=√((2⋅19.6)/9.8)=2s. The path covered by the body is S_2=h-(g(t_0-t_2 )^2)/2=19.6-(9.8⋅〖1.9〗^2)/2=1.911m. Answer: S_1=0.049m,S_2=1.911m.
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