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Question-11536




Question Number 11536 by tawa last updated on 27/Mar/17
Answered by ridwan balatif last updated on 28/Mar/17
Data: m=0.140kg,  v_o =25m/s  (a) EK=(1/2)mv_o ^2                    =(1/2)×0.140×25^2                    =43.75J  (b)if the ball wanna reach maximum height          it′s velocity at maximum height must be zero (v_t =0)  v_t ^2 =v_o ^2 −2gh  0^2 =25^2 −2×10×h  h =32.5 m  (c)v^2 =v_o ^2 −2gh         v^2 =25^2 −2×10×30           v=5m/s  EK=(1/2)mv^2           =(1/2)×0.14×5^2           =1.75J
Data:m=0.140kg,vo=25m/s(a)EK=12mvo2=12×0.140×252=43.75J(b)iftheballwannareachmaximumheightitsvelocityatmaximumheightmustbezero(vt=0)vt2=vo22gh02=2522×10×hh=32.5m(c)v2=vo22ghv2=2522×10×30v=5m/sEK=12mv2=12×0.14×52=1.75J
Commented by tawa last updated on 28/Mar/17
I really appreciate sir. God bless you.
Ireallyappreciatesir.Godblessyou.
Answered by Nysiroke last updated on 28/Mar/17
where h=?  v^2  =u^2  −2gh(upward is−ve)  given that u=25ms^(−1)  ,v=0,g=10ms^(−2)   so  0^2  =25^2  −2×10×h  0=625−20h  ∵20h=625  h=((625)/(20))    =125/4=31.25m(max.height reached by the ball)    a) total energy of the ball=K.E+mgh          =(1/2)mu^2  +mgh          =(m/2)(u^2  +gh),where h=31.25m        =((0.140)/2)(25^2  +10×31.25)        =0.070(625+312.5)         =0.070(937.5)         =65.63J  ci)velocity at 30m upward is given by  v^2 =2gh  v^2 =2gh   =2×10×30 =600  ∴v=(√(600))        =24.50ms^(−1)   cii) k.e at 30m is given by   KE=(1/2)mv^2            =(1/2)×0.140×(√(600^2 ))            =0.070×600           =42J
whereh=?v2=u22gh(upwardisve)giventhatu=25ms1,v=0,g=10ms2so02=2522×10×h0=62520h20h=625h=62520=125/4=31.25m(max.heightreachedbytheball)a)totalenergyoftheball=K.E+mgh=12mu2+mgh=m2(u2+gh),whereh=31.25m=0.1402(252+10×31.25)=0.070(625+312.5)=0.070(937.5)=65.63Jci)velocityat30mupwardisgivenbyv2=2ghv2=2gh=2×10×30=600v=600=24.50ms1cii)k.eat30misgivenbyKE=12mv2=12×0.140×6002=0.070×600=42J
Commented by tawa last updated on 28/Mar/17
God bless you sir.
Godblessyousir.

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