For a projectile, the ratio of maximum height reached to the square of flight time is ($g = 10 ms^{-2}$)
For a projectile, the ratio of maximum height reached to the square of flight time is ($g = 10 ms^{-2}$)
$5:4$
$5:2$
$5:1$
$10:1$
For a projectile, the ratio of maximum height reached to the square of flight time is ($g = 10 ms^{-2}$)
$H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ and $T = \frac{{2u\sin \theta }}{g}$
So $\frac{H}{{{T^2}}} = \frac{{{u^2}{{\sin }^2}\theta /2g}}{{4{u^2}{{\sin }^2}\theta /{g^2}}} = \frac{g}{8} = \frac{5}{4}$