A diver running 1.8 m/s dives out horizontally from the edge of the cliff and 3 s later he reaches the water below.

(a) Calculate the resultant velocity (with direction) of the diver just before he hits the water. (10)
(b) How high is the cliff, and how far from its base did the diver hit the water?

v = √(1.8^2 + (9.8*3)^2) = 29.45

If θ is the angle from the vertical, tanθ = 1.8/(3*9.8)

h = 4.9*3^2 = 44.1 m
distance from base = 3*1.8 = 5.4 m

(a) Well, let's dive into the calculations! Since the diver is moving horizontally, the horizontal component of his velocity remains constant at 1.8 m/s. Just before hitting the water, since he's been falling for 3 seconds, we can say that the initial vertical velocity (when he first jumped) is 0, and the final vertical velocity (just before hitting the water) is what we need to find.

Using the equation of motion, vf = vi + at, we can find the final vertical velocity:

vf = vi + at
vf = 0 + (-9.8 m/s^2)(3 s)
vf = -29.4 m/s

So, the resultant velocity just before he hits the water is a combination of the horizontal and vertical velocities. We can use the Pythagorean theorem to find it:

resultant velocity = √(horizontal velocity^2 + vertical velocity^2)
resultant velocity = √(1.8 m/s)^2 + (-29.4 m/s)^2)
(resultant velocity ≈ 29.5 m/s)

Now, since the diver is falling downwards, let's assume the resultant velocity is negative, which means the direction of the resultant velocity is downwards.

(b) To find the height of the cliff, we can use the formula for vertical displacement:

Δy = viy * t + 0.5 * a * t^2

Since the initial vertical velocity (viy) is 0, we can simplify the equation to:

Δy = 0.5 * a * t^2
Δy = 0.5 * (-9.8 m/s^2) * (3 s)^2
Δy ≈ -44.1 m

Since we know that the diver is falling downwards, we assume the height of the cliff is positive. Therefore, the height of the cliff is approximately 44.1 meters.

To find the horizontal distance from the base of the cliff where the diver hits the water, we can use the formula for horizontal velocity:

Δx = vix * t

Since the initial horizontal velocity (vix) is 1.8 m/s, and the time (t) is 3 seconds, we can calculate:

Δx = 1.8 m/s * 3 s
Δx ≈ 5.4 m

So, the diver hits the water approximately 5.4 meters from the base of the cliff. Now, that's what I call a "cliff-hanger"!

To solve these problems, we can use the equations of motion. Let's assume the initial position of the diver is at the edge of the cliff, and the positive direction is downwards.

(a) To calculate the resultant velocity of the diver just before hitting the water, we need to find the final velocity.
Using the equation: vf = vi + at
where
vf = final velocity
vi = initial velocity (horizontal velocity of 1.8 m/s)
a = acceleration (we assume it to be 0 since the diver is moving horizontally)
t = time (3 seconds)

Substituting the known values into the equation:
vf = 1.8 m/s + 0 m/s^2 * 3 s
vf = 1.8 m/s

Therefore, the resultant velocity of the diver just before hitting the water is 1.8 m/s, in the horizontal direction.

(b) To find the height of the cliff and the distance from its base where the diver hits the water, we need to use the equations of motion.

The equation to find the height of the cliff is:
h = vi * t + 0.5 * a * t^2
where
h = height of the cliff
vi = initial vertical velocity (we assume it to be 0 since the diver is diving horizontally)
a = acceleration due to gravity (-9.8 m/s^2, considering the positive direction as upwards)
t = time (3 seconds)

Substituting the values into the equation:
h = 0 m/s * 3 s + 0.5 * (-9.8 m/s^2) * (3 s)^2
h = -44.1 m/s^2 * 3 s^2 = -44.1 * 9 m

Therefore, the height of the cliff is 44.1 meters.

To find the distance from the base of the cliff, we can use the equation:
d = vi * t
where
d = distance from the base of the cliff (which is the same as the horizontal distance the diver traveled)
vi = initial horizontal velocity (1.8 m/s)
t = time (3 seconds)

Substituting the values into the equation:
d = 1.8 m/s * 3 s = 5.4 m

Therefore, the diver hit the water at a distance of 5.4 meters from the base of the cliff.

To answer part (a), we need to calculate the resultant velocity of the diver just before he hits the water.

The horizontal velocity of the diver remains constant at 1.8 m/s throughout the dive because there are no horizontal forces acting on the diver.

We can use the equation of motion to calculate the vertical velocity of the diver just before he hits the water. The equation for vertical velocity is:

v = u + gt

where:
v = final velocity (unknown)
u = initial velocity (0 m/s, as the diver starts from rest vertically)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken (3 s)

Substituting the known values into the equation:

v = 0 + (9.8 m/s^2)(3 s)
v = 29.4 m/s

Therefore, the resultant velocity of the diver just before he hits the water is 29.4 m/s (downward direction).

To answer part (b), we need to calculate the height of the cliff and the distance from its base where the diver hits the water.

First, let's calculate the vertical displacement of the diver using the equation for displacement:

s = ut + (1/2)gt^2

where:
s = displacement (unknown)
u = initial velocity (0 m/s)
t = time taken (3 s)
g = acceleration due to gravity (9.8 m/s^2)

Substituting the known values into the equation:

s = 0 + (1/2)(9.8 m/s^2)(3 s)^2
s = 0 + (1/2)(9.8 m/s^2)(9 s^2)
s = 0 + (1/2)(88.2)
s = 44.1 m

Therefore, the height of the cliff is 44.1 meters.

To find the horizontal distance from the base of the cliff where the diver hits the water, we can use the horizontal velocity and the time taken:

distance = velocity × time

distance = 1.8 m/s × 3 s
distance = 5.4 m

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