A projectile is fired from the surface of the earth with an initial...

Let the initial speed be $u$ and the angle of projection with the horizontal be $\theta$.

At the highest point of the trajectory, the vertical component of velocity becomes zero, and the projectile only has a horizontal component of velocity, which remains constant throughout the motion. Thus, the speed at the highest point is $v = u \cos\theta$.

Given that the speed at the highest point is half of the initial speed:
$u \cos\theta = \dfrac{u}{2}$
$\cos\theta = \dfrac{1}{2} \implies \theta = 60^{\circ}$

The maximum height $H$ is given by the formula:
$H = \dfrac{u^2 \sin^2\theta}{2g}$

Substituting $u = 40 \mathrm{~m/s}$, $\theta = 60^{\circ}$, and $g = 10 \mathrm{~m/s²}$:
$H = \dfrac{(40)^2 \times (\sin 60^{\circ})^2}{2 \times 10}$
$H = \dfrac{1600 \times \left(\dfrac{\sqrt{3}}{2}\right)^2}{20}$
$H = \dfrac{1600 \times \dfrac{3}{4}}{20} = \dfrac{1200}{20} = 60 \mathrm{~m}$

Answer: $60 \mathrm{~m}$