JEE Advanced Model Paper 6 with solutions for online practice

If v0 is the velocity of projection and α the angle of projection, the equation of trajectory is
y = xtanα -

1

2

gx2

v02cos2α

..... (1)
With origin at the point of projection,
gx2 - 2v02 sinα cosαx + 2v02cos2αy = 0 ..... (2)
Since the projectile passes through two points a,h and 2a,h , then a and 2a must be rootsof equation (2)
a + 2a =

2v02sinαcosα

g

..... (3)
And a × 2a =

2v02cos2αh

g

..... (4)
Dividing eqns. (3) by (4), we get

3a

2a2

=

tanα

h

or tan α =

3h

2a

From eqn (4) v02 =

ga2

h

sec2α =

ga2

h

(1 + tan2α)
=

ga2

h

(1+

9h2

4a2

) =

g

4

(

4a2

h

+9h) or v0 =

1

2

√(

4a2

h

+9h)g