Thank you for registering.

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

Please check your email for login details.
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

The ratio of the magnitudes of the angular momentum of a projectile, at the highest point on trajectory and that at the point where it strikes the ground, about the point of projection is?

The ratio of the magnitudes of the angular momentum of a projectile, at the highest point on trajectory and that at the point where it strikes the ground, about the point of projection is?

Grade:11

2 Answers

Chetan Mandayam Nayakar
312 Points
10 years ago

Dear Varsha,

 

m((u2sin2(theta)/2g)*ucos(theta))/m((2u2sin(theta)cos(theta)/g)*usin(theta)=(1/2)/2=1/4

 

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

 

Askiitians Expert

Chetan Mandayam Nayakar – IIT Madras

vikas askiitian expert
509 Points
10 years ago

angular momentam = mr*v

at highest point V = vcos@(i) &  r = R/2(i) + H(j)             

R is range & H is max height of projectile..

angular momentam=m(r*v)=mvcos@H (-k) ...............1

when the particle strikes the ground then its velocity v = vcos@(i)+vsin@(-j)

 r = R(i) + 0(j)

angular momentam=m(r*v)

                           =mvRsin@(-k)............2

ratio = eq1/eq2=cos@H/sin@R=1/4

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free