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

We are told that the maximum height reached by a projectile is 64 m, and we need to find the new maximum height when the initial velocity is halved. Since the maximum height depends on the vertical component of motion, we recall that for a projectile launched with initial velocity $$u$$ at angle $$\theta$$ with the horizontal, the maximum height is given by $$H = \frac{u^2 \sin^2\theta}{2g}$$. This formula follows from the condition that at maximum height the vertical velocity becomes zero; applying $$v^2 = u_y^2 - 2gH$$ with $$v = 0$$ and $$u_y = u\sin\theta$$ yields $$0 = u^2\sin^2\theta - 2gH \implies H = \frac{u^2\sin^2\theta}{2g}$$.

From this expression, it is clear that with $$\theta$$ and $$g$$ held constant, the maximum height is proportional to the square of the initial velocity, $$H \propto u^2$$. Therefore, if the initial velocity is halved ($$u' = u/2$$), the new maximum height becomes $$H' = \frac{(u/2)^2 \sin^2\theta}{2g} = \frac{u^2 \sin^2\theta}{4 \times 2g} = \frac{H}{4}$$. Substituting the known value $$H = 64$$ m gives $$H' = \frac{64}{4} = 16 \text{ m}$$.

The answer is 16 m.