1. Use the formula for maximum height of a projectile

For a projectile launched with speed $u$ at angle $\theta$,

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

Let the maximum heights of the two balls be $H_1$ and $H_2$.

Given:

$H_1 = 8H_2$

Since both balls have the same initial speed $u$,

$\frac{H_1}{H_2} = \frac{u^2 \sin^2\theta_1}{u^2 \sin^2\theta_2} = \frac{\sin^2\theta_1}{\sin^2\theta_2} = 8$

So,

$\frac{\sin\theta_1}{\sin\theta_2} = \sqrt{8} = 2\sqrt{2}$

2. Use the formula for time of flight

For a projectile,

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

Therefore,

$\frac{T_1}{T_2} = \frac{2u\sin\theta_1/g}{2u\sin\theta_2/g} = \frac{\sin\theta_1}{\sin\theta_2}$

Using the result above,

$\frac{T_1}{T_2} = 2\sqrt{2}$

Hence,

$T_1 : T_2 = 2\sqrt{2} : 1$

3. Check options

• A: $2:1$ ❌
• B: $\sqrt{2}:1$ ❌
• C: $2\sqrt{2}:1$ ✅
• D: $4:1$ ❌

So the correct option is C.