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
Menu
Grade: 12th pass
        a PROJECTILE is thrown at an angle 30° with initial velocity 50 cm/s .when it is at the highest point it explodes and breaks into ratio 1:3 .the smaller part falls down vertically .find the distance where the heavier part lands.
one year ago

Answers : (1)

Vikas TU
6869 Points
							
From momentum conservation in x and y direcn. separetely we get,
In x direcn.
Mvcos30 = M/4*0 + (3M/4)v’
v’ = 2*v*root(3)/3 m/s
In y direcn.
Mvsin30 = + v0*3M/4
v0 = 2v/3
Resultant velocity of heavier mass =>
=> root((2v/3)^2 + (2*v*root(3)/3)^2)
=> root(4v^2/9 + 4v^2/3)
=> root(16v^2/9)
=> 4v/3 m/s
Distance travelled,
(4v/3)^2 = v^2 + 2gs
we get,
s = 7v^2/18g
  = 7*0.25/180
  = 9.72222 x 10^-3 meter.
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 276 off

COUPON CODE: SELF10


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Get Extra Rs. 297 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details