Please please solve question 11
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h = 1/2gt^2
t = √45*2/10 = √9.0 = 3
now , for2 case
h = ut + 1/2gt^2
45 = u*(3-1) + 1/2*10*(3-1)^2
u = 45 - 5*(3)^2/3
u = 45 - 15/3
u = 12.5 .
When a body is reached to the ground from the height 'h' then the initial velocity (u) of body becomes zero.
[vice versa]
In case 1 velocity (initial) of
body becomes zero ( and both balls reach ground at same time.So, we taking a condition first and time be 't sec ')
using second formulae of motion
u = 0 m/sec
t = ?
a = g = 10 m/sec
S = 45 m
Now we get ball takes ''3 sec'' to reach the ground and in question we have given that the second ball dropped after 1 sec of second ball then it's time is (3-1) sec = (2 sec)
S = 45 m
t = (3s-1s) = 2s
a = g = 10 m/sec
u = ?