#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 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)6.75Jb) 67.5Jc)135Jd)1350JPlz explain..

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)