badge image

Enroll For Free Now & Improve Your Performance.

User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

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

Please check your email for login details.

Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 12


What is the relation between Kinetic Energy and Max.Heights? How to solve these kind of problems;- Ratio of min.Kinetic energies of two projectiles of same mass is 4:1.REatio of Max.height attained by them is also 4:1.Ratio of their ranges would be? a)16:1 b)4:1 c)8:1 d)2:1 Plz explain this elaborately.

11 years ago

Answers : (3)

Pratham Ashish
17 Points
    if we have a partical with velocity v then 
                            kinetic energy, K = 1/2 m v^2 
                           max height attained by this ,H = v^2 / 2 g
     bt inthis case velocity must be vertical , otherwise we have to use its vertical component , v sin x  instead of it in the H expression
                       so , we can easily get , k = mg H ,

problem solution,
         let the velocities of two particals be v1, v2
                   angle with horizontal be x1 ,x2
                     kinetic energies  k1, k2
                      max. height     H1, H2 
                  & range r1 ,r2 

   k1/k2 = 4 /1 , gives
    v1 /v2 = 2/1
we know  that,
  H1 =  (v1sinx1 )^2 /2g
   H2 =  (v2sinx2 )^2 /2g

  so H1 / H2 = 4/ 1
gives  v1sinx1 / v2sinx2  =  2/ 1
 from eq (1),.............
 we get sinx1 = sinx2 
 so, x1 = x2

range of first r1 =  2v1 sinx cosx / g
          second, r2 = 2v2 sinx cosx / g

so, r1/ r2 = v1/ v2 = 2/1

ans is d option.

11 years ago
Pavan kumar
18 Points

The kinetic energies are minimum at the maximum height where velocity is ucosx

Hence 1/2 mu^2cos^2x1= 4*1/2 mv^2cos^2x2

or, ucosx1 = 2vcosx2........(1)

Maximum height is given by H = u^2sin^x/2g

Hence u^2sin^2x1/2g = 4*v^2sin^2x2/2g

or, usinx1 = 2vsinx2.......(2)

(2)/(1) gives tanx1=tanx2 or x1=x2

Using in (1) we get ucosx1 = 2vcosx1

hence u=2v

Therefore R1/R2 = {u^2sin2x1/g}/{v^2sin2x2/g} = (2v)^2sin2x1/v^2sin2x2  = 4

Hence option B is correct.

11 years ago
11 Points
hope it might help you
thank you
3 years 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
To Win!!! Click Here for details