Click to Chat

1800-2000-838

+91-120-4616500

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
`        two particle move in uniform gravitational field with an accelration 'g'.At the initial moment ,the particle were located over a tower at one point and moves with velocity v1=3m/s and v2=4m/s horizontally in opposite direction. find distance between them at the moment when thier velocity vector become mutually perpendicular?`
7 years ago

419 Points
```										dear Rajan
see the pic for solution
All the best.
AKASH GOYAL

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
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.

```
7 years ago
510 Points
```										at any time t
velocity of first particle is V1 =3i - gt j
position of first paticle is S1= 3ti -gt^2/2 j

velocity of second particle is V2= -4i -gt j
position of second particle is S2 =-4t -gt^2/2 j                                     (i and j are unit vectors along x axis and y axis)
if their velocities are mutually perpendicular then dot  product will will be 0...
V1.V2=0
t^2 = 3/25 units
distance bw them at dis time is S1-S2=5t= (7sqrt3)/5 units
```
7 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details