For a projectile, the ratio of maximum height reached to the square of flight time is (g = 10 ms⁻²)

The correct option is 5 : 4.

To find the ratio of the maximum height reached by a projectile to the square of its time of flight, we start by writing down the standard formulas for both quantities.

Let a projectile be launched with an initial velocity u at an angle θ with the horizontal. The acceleration due to gravity is represented by g.

The maximum height (H) reached by the projectile is given by the formula:

$H=\frac{u^2{\sin }^2\theta }{2g}$

The total time of flight (T) of the projectile is given by the formula:

$T=\frac{2u\sin \theta }{g}$

Squaring the time of flight gives:

$T^2=\frac{4u^2{\sin }^2\theta }{g^2}$

Now, we find the ratio of maximum height to the square of the time of flight:

$\frac{H}{T^2}=\frac{\left(\frac{u^2{\sin }^2\theta }{2g}\right)}{\left(\frac{4u^2{\sin }^2\theta }{g^2}\right)}$

We can simplify this expression by canceling the common terms, $u^2{\sin }^2\theta$, in both the numerator and the denominator:

$\frac{H}{T^2}=\frac{1}{2g}\times \frac{g^2}{4}$

Simplifying further gives:

$\frac{H}{T^2}=\frac{g}{8}$

Given that the acceleration due to gravity $g=10{\text{ms}}^{-2}$, we substitute this value into the equation:

$\frac{H}{T^2}=\frac{10}{8}=\frac{5}{4}$

Therefore, the ratio of maximum height to the square of flight time is 5 : 4.