Answer 1

Given:
Height of the cliff, h = 40 m 
Initial speed of the projectile, u = 50 m/s
Let the projectile hit the ground with velocity 'v'.
Applying the law of conservation of energy,

\[\text{ mgh } + \frac{1}{2}\text{mu}^2 = \frac{1}{2}\text{mv}^2 \]

\[ \Rightarrow 10 \times 40 + \left( \frac{1}{2} \right) \times 2500 = \frac{1}{2} \text{v}^2 \]

\[ \Rightarrow \text{v}^2 = 3300\]

\[ \Rightarrow \text{ v = 57 . 4 m/s = 58 m/s} \]

The projectile hits the ground with a speed of 58 m/s.