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A body of mass 0.2kg is projected obliquely from a point on horizontal ground with a certain velocity.
At two Points A&B in its path,its velocitys are found to be 30m/s and 30(root3)m/s respectivly.
If these velociteis are perpendicular to each other, max. K.E of the body is:
a)6.75J
b) 67.5J
c)135J
d)1350J
Plz explain..
(Vh)^2 +(Vv1)^2=900
(Vh)^2 +(Vv2)^2=2700
Vh is horizontal component of velocity, and Vv is vertical component
for perpendicularity,dot product is zero.
(Vhi+Vv1j)*(Vhi-Vv2j)=0
Vh^2=Vv1*Vv2,
find Vh,Vv1,Vv2 from the above equations
let theta be angle of projection
Vv1^2=Vv^2-2gh1
Vv2^2=Vv^2-2gh2,
(Vv2)^2-Vv1^2=2g(h1-h2)
answer=(m/2)(Vh^2+Vv^2)
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