Here, the trajectory of a projectile motion,
y=αx−βx2 .......(i)
As, general equation of trajectory of a projectile motion,
y=tanθ.x−.x2 ........(ii)
Now, comparing Eqs. (i) and (ii), we get
tanθ=α⇒=α2 ........(iii)
and
β= ⇒4β= ........(iv)
Maximum height of projectile,
Hmax= From Eqs. (iii) and (iv), we get
==Hmax Hence, the maximum height is
.