Answer:
Approximately [tex]11\; {\rm m}[/tex], assuming that gravitational acceleration is [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] and that air resistance is negligible.
Explanation:
The question is asking for the vertical displacement, [tex]x_{y}[/tex].
Under the assumptions, acceleration in the vertical direction would be [tex]a_{y} = -g = (-9.81)\; {\rm m\cdot s^{-2}}[/tex].
Since the projectile is launched horizontally, initial vertical velocity would be zero: [tex]u_{y} = 0\; {\rm m\cdot s^{-1}}[/tex].
It is given that the projectile is in this motion for [tex]t = 2.2\; {\rm s}[/tex]. Apply the SUVAT equation [tex]x = (1/2)\, a\, t^{2} + u\, t[/tex] and solve for the vertical displacement:
[tex]\begin{aligned}x_{y} &= \frac{1}{2}\, a_{y}\, t^{2} + u_{y}\, t \\ &= \frac{1}{2}\, ((-9.81)\; {\rm m\cdot s^{-2}})\, (2.2\; {\rm s})^{2} + (0\; {\rm m\cdot s^{-1}})\, (2.2\; {\rm s}) \\ &= \frac{(-9.81)\, (2.2)^{2}}{2}\; {\rm m} \\ &\approx (-11)\; {\rm m} \end{aligned}[/tex].
Note that the value of vertical displacement is negative since the projectile landed below where it was launched. The height of this cliff would be approximately [tex]11\; {\rm m}[/tex].
