Answer:
Approximately [tex]63.6\; \rm m[/tex] assuming that air resistance on this projectile is negligible.
Explanation:
Assume that the air resistance on this projectile is negligible. The projectile would accelerate downwards at [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].
Since the acceleration of this projectile is vertical, the horizontal velocity of this projectile would be constant. The projectile would continue to travel horizontally at [tex]22.0\; \rm m\cdot s^{-1}[/tex] until it lands.
Apply the SUVAT equations to find the duration [tex]t[/tex] of the flight of this projectile. Since vertical acceleration is constantly [tex]g[/tex], the total vertical displacement of the projectile during the flight would be:
[tex]\displaystyle x = \frac{1}{2}\, g\, t^{2}[/tex].
The vertical displacement of this projectile is equal to the height of this cliff. That is: [tex]x = 41.0\; \rm m[/tex]. Given that [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex], rearrange the equation and solve for the duration of this flight, [tex]t[/tex] ([tex]t > 0[/tex].)
[tex]\begin{aligned}t &= \sqrt{\frac{2\, x}{g}} \\ &= \sqrt{\frac{41.0\; \rm m}{9.81\; \rm m\cdot s^{-2}}}\\ &\approx 2.8926\; \rm s\end{aligned}[/tex].
Since the horizontal velocity of this projectile is constantly [tex]22.0\; \rm m\cdot s^{-1}[/tex], the horizontal displacement of this projectile would be:
[tex]\begin{aligned}& x(\text{horizontal}) \\ =\; & v(\text{horizontal}) \, t\\ =\; & 22.0\; \rm m\cdot s^{-1} \times 2.8926\; \rm s \\ =\; & 63.6\; \rm m\end{aligned}[/tex].
