Answer:
The monkey hit the pond after approximately [tex]0.55\; {\rm s}[/tex] in the air.
The monkey hit the pond approximately [tex]1.1\; {\rm m}[/tex] away from the base of the cliff.
(Assume that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex], air resistance on the monkey is negligible, and the cliff is vertical.)
Explanation:
Assume that the air resistance on the monkey is negligible. The vertical acceleration of the monkey would be constantly [tex]a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex]. Note that [tex]a_{y}[/tex] is negative since the monkey is accelerating downwards.
Right before hitting the pond, the monkey would be [tex]1.5\; {\rm m}[/tex] below the cliff. Hence, the vertical displacement [tex]x_{y}[/tex] of the monkey would be [tex](-1.5)\; {\rm m}[/tex].
Let [tex]u_{y}[/tex] denote the initial vertical velocity of the monkey. Since the top of the cliff is level, initial velocity will be entirely horizontal. Hence, [tex]u_{y} = 0\; {\rm m\cdot s^{-1}}[/tex].
Let [tex]t[/tex] denote the amount of time it took for the monkey to hit the pond ([tex]t \ge 0[/tex].) Rearrange the SUVAT equation [tex]x_{y} = (1/2)\, a_{y}\, t^{2} + u_{y}\, t[/tex] and solve for [tex]t\![/tex].
Since [tex]u_{y} = 0\; {\rm m\cdot s^{-1}}[/tex], the equation simplifies to:
[tex]x_{y} = (1/2)\, a_{y}\, t^{2}[/tex].
[tex]\begin{aligned}t^{2} = \frac{2\, x_{y}}{a_{y}}\end{aligned}[/tex].
Since [tex]t \ge 0[/tex]:
[tex]\begin{aligned}t &= \sqrt{\frac{2\, x_{y}}{a_{y}}} \\ &= \sqrt{\frac{(2)\, (1.5\; {\rm m})}{(9.81\; {\rm m\cdot s^{-2}})}} \\ &\approx 0.553\; {\rm s}\end{aligned}[/tex].
Hence, it would take approximately [tex]0.55\; {\rm s}[/tex] for the monkey to hit the pond.
Also because air resistance on the monkey is negligible, the horizontal velocity [tex]v_{x}[/tex] of the monkey will be constantly equal to the initial value [tex]2.0\; {\rm m\cdot s^{-1}}[/tex].
Hence, the monkey would have travelled a horizontal distance [tex]x_{x}[/tex] of:
[tex]\begin{aligned} x_{x} &= v_{x}\, t \\ &\approx (2.0\; {\rm m\cdot s^{-1}})\, (0.553\; {\rm s})\\ &\approx 1.1\; {\rm m}\end{aligned}[/tex].