UPSEE 2019 Solved Paper

Here, the trajectory of a projectile motion,
y=αx−βx2 .......(i)
As, general equation of trajectory of a projectile motion,
y=tan‌θ.x−

g

2(u‌cos‌θ)2

.x2 ........(ii)
Now, comparing Eqs. (i) and (ii), we get
tan‌θ=α⇒

sin2θ

cos2θ

=α2 ........(iii)
andβ=

g

2u2cos2θ

⇒4β=

2g

u2cos2θ

........(iv)
Maximum height of projectile,
Hmax=

u2sin2θ

2g

From Eqs. (iii) and (iv), we get

α2

4β

=

u2cos2θ×sin2θ

2gcos2θ

=Hmax
Hence, the maximum height is

α2

4β

.