The maximum height reached by a projectile is $64 \mathrm{~m}$. If the initial velocity is halved, the new maximum height of the projectile is ______ $\mathrm{m}$.

<p>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 $g$: <p>$H = \frac{v_0^2 \sin^2(\theta)}{2g}$</p> <p>where:</p> <ul> <li>$H$ is the maximum height,</li><br> <li>$v_0$ is the initial velocity of the projectile,</li><br> <li>$\theta$ is the angle of projection with the horizontal, and</li><br> <li>$g$ is the acceleration due to gravity, which is approximately $9.8 \, \mathrm{m/s^2}$.</li> </ul> <p>Given that the maximum height attained by the projectile is $64 \, \mathrm{m}$, we can write:</p> <p>$64 = \frac{v_0^2 \sin^2(\theta)}{2 \cdot 9.8}$</p> <p>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:</p> <p>$H' = \frac{{v'_0}^2 \sin^2(\theta)}{2g}$</p> <p>Substituting $v'_0 = \frac{v_0}{2}$ into this equation:</p> <p>$H' = \frac{(\frac{v_0}{2})^2 \sin^2(\theta)}{2g} = \frac{v_0^2 \sin^2(\theta)}{2 \cdot 4g} = \frac{1}{4} \cdot \frac{v_0^2 \sin^2(\theta)}{2g}$</p> <p>Since we know the original height:</p> <p>$64 = \frac{v_0^2 \sin^2(\theta)}{2 \cdot 9.8}$</p> <p>Substituting this value into our equation for $H'$, we find:</p> <p>$H' = \frac{1}{4} \cdot 64 = 16 \, \mathrm{m}$</p> <p>Therefore, if the initial velocity of the projectile is halved, the new maximum height reached by the projectile would be $16 \, \mathrm{m}$.</p></p>