1. For a projectile, the maximum height is given by

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

where:

• $u$ = initial speed
• $\theta$ = angle of projection
• $g$ = acceleration due to gravity

2. Initially, the maximum height is

$H_1 = 64\,\text{m}$

So,

$H_1 = \frac{u^2 \sin^2\theta}{2g} = 64$

3. If the initial velocity is halved, the new speed becomes

$u' = \frac{u}{2}$

Then the new maximum height is

$H_2 = \frac{(u')^2 \sin^2\theta}{2g}$

Substitute $u' = \frac{u}{2}$:

$H_2 = \frac{\left(\frac{u}{2}\right)^2 \sin^2\theta}{2g}$

$H_2 = \frac{u^2}{4} \cdot \frac{\sin^2\theta}{2g}$

$H_2 = \frac{1}{4} \left(\frac{u^2 \sin^2\theta}{2g}\right)$

$H_2 = \frac{1}{4} H_1$

4. Therefore,

$H_2 = \frac{1}{4} \times 64 = 16\,\text{m}$

So the new maximum height is:

$\boxed{16}$