To solve this problem, we first need to understand the formula that relates the maximum height $H$ reached by a projectile to its initial velocity $v_0$ and the acceleration due to gravity $g$:

$H = \frac{v_0^2 \sin^2(\theta)}{2g}$

where:

• $H$ is the maximum height,


• $v_0$ is the initial velocity of the projectile,


• $\theta$ is the angle of projection with the horizontal, and


• $g$ is the acceleration due to gravity, which is approximately $9.8 \, \mathrm{m/s^2}$.

Given that the maximum height attained by the projectile is $64 \, \mathrm{m}$, we can write:

$64 = \frac{v_0^2 \sin^2(\theta)}{2 \cdot 9.8}$

Now, we are asked to find the new maximum height if the initial velocity is halved. Let's denote the new initial velocity as $v'_0 = \frac{v_0}{2}$. Using the formula for maximum height again, we get:

$H' = \frac{{v'_0}^2 \sin^2(\theta)}{2g}$

Substituting $v'_0 = \frac{v_0}{2}$ into this equation:

$H' = \frac{(\frac{v_0}{2})^2 \sin^2(\theta)}{2g} = \frac{v_0^2 \sin^2(\theta)}{2 \cdot 4g} = \frac{1}{4} \cdot \frac{v_0^2 \sin^2(\theta)}{2g}$

Since we know the original height:

$64 = \frac{v_0^2 \sin^2(\theta)}{2 \cdot 9.8}$

Substituting this value into our equation for $H'$, we find:

$H' = \frac{1}{4} \cdot 64 = 16 \, \mathrm{m}$

Therefore, if the initial velocity of the projectile is halved, the new maximum height reached by the projectile would be $16 \, \mathrm{m}$.