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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..

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..

Grade:12

1 Answers

Chetan Mandayam Nayakar
312 Points
10 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)

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