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 touches 4 verices of a regular hexagon of side length 10√3. Find its time of flight.
one month ago

Answers : (1)

Arun
22031 Points
							
 

let projectile is projected at a distance a from the corner of the base so its range must be

 R = a+10√3 + a

      = 2a +10√3

so 2a + 10√3 = V2 sin2b/2g  

 or  g/v2cos2b = sinb/(2a+10√3)  ....................1

select first two corner point of hexagon ,if the projectile pass through these point then by symmetry it will also pass through oteher two ppoints

 points are ( a-10√3cos60 , 10√3sin60)   and  (a,20√3sin60)

use the general equation

y =x tanb - gx2/2v2cos2b

put above these two point and put the value of g/v2cos2b = sinb/(2a+10√3) 

then u will get 2 equation in a and tanb  

solve it  and then easly u can get desired result

one month ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


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


Ask Experts

Have any Question? Ask Experts

Post Question

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