A toy rocket is launched from the ground with an initial speed of...

Given: initial speed $u = 40\text{ m/s}$, maximum height $H = 40\text{ m}$, and $g = 10\text{ m/s}^2$.

The formula for the maximum height of a projectile is:
$H = \dfrac{u^2 \sin^2 \theta}{2g}$

Substitute the known values into the equation:
$40 = \dfrac{(40)^2 \sin^2 \theta}{2 \times 10}$
$40 = \dfrac{1600 \sin^2 \theta}{20}$
$40 = 80 \sin^2 \theta$

Solving for $\sin^2 \theta$:
$\sin^2 \theta = \dfrac{40}{80} = \dfrac{1}{2}$

Taking the square root of both sides:
$\sin \theta = \dfrac{1}{\sqrt{2}}$

The angle whose sine is $\dfrac{1}{\sqrt{2}}$ is $45^{\circ}$. Thus, the angle of projection is $45^{\circ}$.

Answer: $45^{\circ}$