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a bodyprojected up reaches a point A in its path at the end of 4 th second and reaches the ground after 5 seconds from the start the height of A above the the ground is
Time of ascent=Time of descent=2.5s as total time is 5sLet x be the height of point AAt max. height, v=0. Thus,v=u+at0=u+(-g)tu=gt and thus u=(10)2.5= 25m/sWKT, u2=2gHH=25(25)/20=31.25mWe can infer that point A is in the path of descent.Now starting from the max height H, u=0. Already 2.5 seconds are over. So to reach A the body must travel 1.5 seconds more. If H-x is the distance of A from max height,H-x=1/2gt2H-x=1/2(10)(1.5)2H-x=2.25(5)=11.25x=31.25-11.25Thus, x=20mFinal answer=20m
from the aboe question we haveTime of ascent=Time of descent=2.5s as total time is 5sLet x be the height of point AAt max. height, v=0. Thus,v=u+at0=u+(-g)tu=gt and thus u=(10)2.5= 25m/sWKT, u2=2gHH=25(25)/20=31.25mWe can infer that point A is in the path of descent.Now starting from the max height H, u=0. Already 2.5 seconds are over. So to reach A the body must travel 1.5 seconds more. If H-x is the distance of A from max height,H-x=1/2gt2H-x=1/2(10)(1.5)2H-x=2.25(5)=11.25x=31.25-11.25Thus, x=20mFinal answer=20m
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