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Thakgalo sakwana
Questions (1)
An object is projected vertically upwards and returns to the point of projection 12s later. Ignore air resistance and determine
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Answers (4)
Vf= vi +at vf= (0) +(-10)(10) vf=-100m/s downward
100KHP/(KHP+NaHO) =100(0.847)/(0.847+19.82) =4.098...%
Sin¤= 1/3 ¤=19.47' Fnet= Fg// - Ff Fnet= mg.sin¤ -Ff =2(9.8)sin19.47-0.35 Fnet= 6.183N a= Fnet/m a= 6.183/2 a=3.091m/s^2 (Vf^2)=(Vi^2) +2as Vf^2= (0)+2(3.091)(3) Vf^2= 18.546 Vf= 4.31m/s downward. ***GOOD LUCK*** For info call me on 0723624826
Vf=Vi +at Vf=0, Vi=?, a=-9.8 remeber time taken to go up is equal to the time taken to go back to the point of projectile.. THEREFORE t=6 Vf=Vi +at (0)=Vi +(-9.8)(6) Vi=58.8m/s ***[OR]*** Vf=Vi +at Vf=-Vi, Vi=?, a=-9.8, t=12 (-Vi)=Vi +(-9.8)(12) -2Vi=
