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# a  bodyprojected up reaches a point A in its path at the end of 4th second and reaches the ground after 5 seconds from the start the height of A above the the ground is

Raveesh
12 Points
5 years ago
Time of ascent=Time of descent=2.5s as total time is 5s
Let x be the height of point A
At max. height, v=0. Thus,
v=u+at
0=u+(-g)t
u=gt and thus u=(10)2.5= 25m/s
WKT, u2=2gH
H=25(25)/20=31.25m
We 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=1/2gt2
H-x=1/2(10)(1.5)2
H-x=2.25(5)=11.25
x=31.25-11.25
Thus, x=20m

sumanth
38 Points
4 years ago
from the aboe question we have
Time of ascent=Time of descent=2.5s as total time is 5s
Let x be the height of point A
At max. height, v=0. Thus,
v=u+at
0=u+(-g)t
u=gt and thus u=(10)2.5= 25m/s
WKT, u2=2gH
H=25(25)/20=31.25m
We 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=1/2gt2
H-x=1/2(10)(1.5)2
H-x=2.25(5)=11.25
x=31.25-11.25
Thus, x=20m