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Grade 12Mechanics

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..

Profile image of Chilukuri Sai Kartik
15 Years agoGrade 12
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1 Answer

Profile image of Chetan Mandayam Nayakar
14 Years ago

(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)