A projectile is launched from the top of a 40 m high cliff with an...

Using the principle of conservation of mechanical energy, the total energy of the projectile remains constant.

Let $m$ be the mass of the projectile. Taking the ground as the reference level for potential energy:

Initial energy $E_i = \dfrac{1}{2}mu^2 + mgh$

Final energy $E_f = \dfrac{1}{2}mv^2$

Equating initial and final energies:

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 + mgh$

$v^2 = u^2 + 2gh$

Substituting the given values ($u = 50 \text{ m/s}$, $h = 40 \text{ m}$) and taking $g = 10 \text{ m/s}^2$:

$v^2 = (50)^2 + 2(10)(40)$

$v^2 = 2500 + 800 = 3300$

$v = \sqrt{3300} \approx 57.4 \text{ m/s}$

The closest value among the given options is $58 \text{ m/s}$.