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Mechanics
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21 Sep 2007 23:02:30 IST
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let the mass of the body be m, and angle of inclination be theta.
ignoring friction,
mg.sin(theta) = m.a
a=g sin(the.)
1/2= 0 + 1/2 .gsin(theta) .(1/2)^2
= sin(theta) = 4/10
= a= 4
v ( after have second) = (2*4*1/2) ^ 1/2
v= 2
now
1/2 = 2t + 2t^2
t= 0.21 seconds
Ignore the negative value in quad. since time cant be -ve.
Plz. rate me!!!
ignoring friction,
mg.sin(theta) = m.a
a=g sin(the.)
1/2= 0 + 1/2 .gsin(theta) .(1/2)^2
= sin(theta) = 4/10
= a= 4
v ( after have second) = (2*4*1/2) ^ 1/2
v= 2
now
1/2 = 2t + 2t^2
t= 0.21 seconds
Ignore the negative value in quad. since time cant be -ve.
Plz. rate me!!!













At t1=1/2, s=1/2 :
1/2 = u(1/2) + (1/2)a(1/2)2
a = 4 (for u=0)
When s=1 :
1 = ut2 + (1/2)at22
t22 = 2/a = 2/4 = 1/2
t2 = 1/
Hence time taken to move next 1/2 metre = t2 - t1 = 1/