KEAM 2013 Physics and Chemistry Paper

From equations of motions, we know
v2=u2−2gh......(i)
Here, v=2v, so
(2v)2=u2+2gh.....(ii)
Now, on adding Eqs. (i) and (ii), we get
5v2=2u2
u2=‌

5

2

v2.....(iii)
On subtracting Eq. (i) in Eq. (ii), we get
3v2=4gh
v2=‌

4

3

gh.....(iv)
And we know that the
H=‌

u2

2g

So, here H=‌

‌ 5 2 v2

2g

=‌

‌ 5 2 ( 4 3 gh)

2g

H=‌

5

3

h