Answer:
(Assumption: air resistance on the rock is negligible; [tex]g = 9.8\; {\rm m\cdot s^{-2}}[/tex].)
The rock travelled [tex]60\; {\rm m}[/tex] horizontally.
The cliff is approximately [tex]20\; {\rm m}[/tex] above the water.
The overall displacement of the rock is approximately [tex]63\; {\rm m}[/tex].
Horizontal velocity of the rock right before it hit the water: [tex]30\; {\rm m\cdot s^{-1}}[/tex] (same as when the rock was launched.)
Vertical velocity of the rock right before it hit the water: approximately [tex]20\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
Assume that the air resistance on the rock is negligible. During the entire flight, the horizontal velocity of the rock will be constantly [tex]v_{x} = 30\; {\rm m\cdot s^{-1}}[/tex]. The vertical velocity of the rock will uniformly increase at a rate of [tex]a_{y} = g = 9.8\; {\rm m\cdot s^{-2}}[/tex] (this rate is the vertical acceleration of the rock.)
Since the horizontal velocity of the rock is constant, multiply that velocity by time to find the horizontal displacement:
[tex]\begin{aligned} x_{x} &= v_{x}\, t \\ &= 30\; {\rm m\cdot s^{-1}} \times 2.0\; {\rm s} \\ &= 60\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Let [tex]u_{y}[/tex] denote the initial vertical velocity of the rock. In this question, [tex]u_{y} = 0\; {\rm m\cdot s^{-1}}[/tex] since the stone was thrown horizontally.
While the vertical velocity of the rock isn't constant, vertical acceleration [tex]a_{y}[/tex] is constant. Hence, apply the SUVAT equation [tex]x_{y} = (1/2)\, a_{y}\, t^{2} + v_{y,0}\, t[/tex] to find the vertical displacement [tex]x_{y}[/tex]:
[tex]\begin{aligned}x_{y} &= \frac{1}{2}\, a_{y}\, t^{2} + v_{y,0}\, t \\ &= \frac{1}{2}\times 9.8\; {\rm m\cdot s^{-2}} \times (2.0\; {\rm s})^{2} + 0\; {\rm m \cdot s^{-1}} \times 2.0\; {\rm s} \\ &= \frac{1}{2}\times 9.8\times 2.0^{2} \; {\rm m} \\ &= 19.6\; {\rm m}\end{aligned}[/tex].
The height of the cliff (relative to the water underneath) is equal to the vertical displacement of the rock: [tex]19.6\; {\rm m}[/tex] (approximately [tex]20\; {\rm m}[/tex] when rounded to two significant figures.)
The horizontal and vertical displacement of this rock are perpendicular to one another. These two displacements [tex]x_{x}[/tex] and [tex]x_{y}[/tex] form the two sides of a right triangle, with the overall displacement [tex]x[/tex] as the other side (the hypotenuse.) The overall displacement of this rock can be found using the Pythagorean Theorem:
[tex]\begin{aligned}x &= \sqrt{{x_{x}}^{2} + {x_{y}}^{2}} \\ &= \sqrt{(60\; {\rm m})^{2} + (19.6\; {\rm m})^{2}} \\ &\approx 63\; {\rm m}\end{aligned}[/tex].
Again, assume that air resistance on the rock is negligible. The horizontal velocity of the rock right before hitting the water will be the same as the initial value: [tex]30\; {\rm m\cdot s^{-1}}[/tex].
Let [tex]v_{y}[/tex] denote the vertical velocity of the rock right before hitting the water. Apply the SUVAT equation [tex]v_{y} = u_{y} + a\, t[/tex] to find [tex]v_{y}\![/tex]:
[tex]\begin{aligned} v_{y} &= u_{y} + a\, t \\ &= 0\; {\rm m\cdot s^{-1}} + 9.8\; {\rm m\cdot s^{-2}} \times 2.0\; {\rm s} \\ &= 19.6\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Hence, right before hitting the water, the vertical velocity of this rock will be [tex]19.6\; {\rm m\cdot s^{-1}}[/tex] (rounded to [tex]20\; {\rm m\cdot s^{-1}}[/tex].)
