A diver running 1.8m/s dives out horizontally from the edge of a vertical cliff and 3.0s later reaches the water below. How high was the cliff, and how far from its base did the diver hit the water?

h = ½gt², since diver's initial vertical speed is 0

h = 44 m

x = vt
x = 5.4 m

Why did the diver jump off the cliff? He wanted to make a splash in his diving career!

Anyway, let's get to the calculations. We'll use the kinematic equations to solve the problem. We have the initial velocity (vi) of the diver, which is 0 m/s since he started from rest on the edge of the cliff. The final velocity (vf) is 1.8 m/s as given in the problem.

Using the equation vf = vi + at, we can solve for acceleration (a). Rearranging the equation, we get a = (vf - vi) / t.

Substituting the values, we get a = (1.8 m/s - 0 m/s) / 3.0 s.

So, a = 1.8 m/s / 3.0 s = 0.6 m/s^2.

Now, we'll use the equation d = vi*t + (1/2)at^2 to find the height of the cliff (d) and the horizontal distance (x) the diver covered.

For the height of the cliff:
d = vi*t + (1/2)at^2,
d = 0 m/s * 3.0 s + (1/2) * (0.6 m/s^2) * (3.0 s)^2,
d = 0 + 0.9 m.

So, the height of the cliff is approximately 0.9 meters.

For the horizontal distance from the base of the cliff:
x = vi*t,
x = 0 m/s * 3.0 s,
x = 0 meters.

That's right, our diver didn't make it very far horizontally. He had a great vertical dive though!

To solve this problem, we can use the equations of motion to find the height of the cliff and the horizontal distance the diver traveled before hitting the water.

Let's first find the horizontal distance the diver traveled. We know that the diver's horizontal velocity is 1.8 m/s, and the time taken to reach the water is 3.0 seconds. We can use the formula:

horizontal distance (d) = velocity (v) x time (t)

Substituting the given values:

d = 1.8 m/s x 3.0 s
d = 5.4 m

So, the diver hit the water 5.4 meters from the base of the cliff horizontally.

Now, let's find the height of the cliff. We can use the equation of motion for vertical motion:

vertical distance (h) = initial velocity in the vertical direction (u) x time (t) + (1/2) x acceleration (a) x time squared (t^2)

Since the diver starts from rest vertically (u = 0 m/s) and falls downwards, we can simplify the equation to:

h = (1/2) x a x t^2

The acceleration due to gravity is approximately 9.8 m/s^2 (assuming no air resistance).

h = (1/2) x 9.8 m/s^2 x (3.0 s)^2
h = 44.1 m

So, the height of the cliff is 44.1 meters.

To summarize:
- The height of the cliff is 44.1 meters.
- The diver hit the water 5.4 meters from the base of the cliff horizontally.

To solve this problem, we can use kinematic equations of motion. Let's break it down into two parts: the horizontal motion and the vertical motion of the diver.

1. Horizontal Motion:
Since the diver moves horizontally at a constant speed of 1.8 m/s, we know the horizontal displacement is given by:
Horizontal displacement = Speed × Time = 1.8 m/s × 3.0 s = 5.4 m

Therefore, the diver hits the water 5.4 meters away from the base of the cliff horizontally.

2. Vertical Motion:
Let's consider the vertical motion of the diver. The equation to determine the height of the cliff can be obtained by using the equation of motion:
Vertical displacement = Initial vertical velocity × Time + (0.5 × Acceleration × Time^2)

We know the final vertical displacement is zero (as the diver lands in the water), initial vertical velocity is zero (the diver starts from rest vertically), and the acceleration due to gravity is approximately -9.8 m/s^2 (taking upward as positive). Let's plug in the values and solve for the initial vertical displacement, which is the height of the cliff.

0 = 0 × 3.0 s + (0.5 × (-9.8 m/s^2) × (3.0 s)^2
0 = -14.7 s^2
s^2 = 0

Therefore, the initial vertical displacement is zero. This means that the cliff has no height in the vertical direction.

Hence, the cliff has zero height vertically, and the diver hits the water 5.4 meters away from the base of the cliff horizontally.