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```        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.
```
8 years ago

Pratham Ashish
17 Points
```										 hi,
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.

```
8 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.
```
8 years ago
hshhasjhj
11 Points
```										gfytuhghghghghgbhjgdhjd 58527kejiufwesmncnsdnjkjfrev 424245  ekjfnds   242442   “””;,.d./w  hope it might help you      thank you
```
9 months ago
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