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

particle is projected vertically upwards from O with velocity v and a second particle projected at same instant from P at height h above O with velocity v and angle of projection theta . find tme after which distance between the 2 s minimum ?

particle is projected vertically upwards from O with velocity v and a second particle projected at same instant from P at height h above O with velocity v and angle of projection theta . find tme after which distance between the 2 s minimum ?

Grade:11

1 Answers

Chetan Mandayam Nayakar
312 Points
9 years ago

position vector of 2nd particle= vtcos(theta)i +(h+(vtsinθ -(g/2)t^2)j,position vector of first particle=vt-(g/2)t^2 j

subtract one position vector from the other and obtain the relative positive vector as a function of t, minimize it with respect to find the answer, vector connecting particles= vtcos(theta)i +(h-vt(1-sin(theta))

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