A diver springs upward with an initial speed of 1.66 m/s from a 3.0-m board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = -3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.] (b) What is the highest point he reaches above the water?

a. V^2 = Vo^2 + 2g*h

h(up) = (V^2-Vo^2)/2g.
h(up) = (0-2.76)/-19.6 = 0.141 m.

V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6(0.141+3) = 61.6
V = 7.85 m/s.

b. 0.141 m.

CORRECTION:

b. hmax = 3 + 0.141 = 3.141 m.

To solve this problem, we can use the equations of motion for uniformly accelerated motion. The equation we will use is:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement

(a) To find the velocity with which the diver strikes the water, we can use this equation with the following values:

Initial velocity (u) = 1.66 m/s (upward)
Displacement (s) = -3.0 m (downward)
Acceleration (a) = -9.8 m/s^2 (downward, due to gravity)

Plugging in these values, we have:

v^2 = (1.66 m/s)^2 + 2(-9.8 m/s^2)(-3.0 m)

v^2 = 2.7556 + 58.8

v^2 = 61.5556

Taking the square root of both sides:

v = √61.5556

v ≈ 7.849 m/s

Therefore, the velocity with which the diver strikes the water is approximately 7.849 m/s.

(b) To find the highest point the diver reaches above the water, we can calculate the distance traveled during the upward motion using the equation:

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

Where:
s = displacement (unknown)
u = initial velocity (1.66 m/s, upward)
a = acceleration (-9.8 m/s^2, downward)
t = time

At the highest point, the final velocity (v) will be zero. So we can use this fact to determine the time it takes to reach the highest point:

v = u + at

0 = 1.66 m/s + (-9.8 m/s^2)t

-1.66 m/s = -9.8 m/s^2t

t = 1.66 m/s / 9.8 m/s^2

t ≈ 0.169 s

Now we can substitute this value of t into the equation for displacement:

s = (1.66 m/s)(0.169 s) + (1/2)(-9.8 m/s^2)(0.169 s)^2

s ≈ 0.279 m

Therefore, the highest point the diver reaches above the water is approximately 0.279 meters.

To find the answers to these questions, we can use the equations of motion for an object in free fall. Let's break down the problem step by step and find the solution.

(a) Find the velocity with which he strikes the water:

We can use the equation of motion:

vf^2 = vi^2 + 2ax

where
vf = final velocity (unknown)
vi = initial velocity (1.66 m/s, positive since it's upward)
a = acceleration due to gravity (-9.8 m/s^2, negative since it's downward)
x = displacement (y = -3.0 m, negative since it's downward)

Plugging in the given values, we can solve for the final velocity:

vf^2 = (1.66 m/s)^2 + 2*(-9.8 m/s^2)*(-3.0 m)
vf^2 = 2.7556 m^2/s^2 - 58.8 m^2/s^2
vf^2 = -56.0444 m^2/s^2

Since velocity can't be negative in this case (as it is a scalar quantity), we take the positive square root:

vf = √(-56.0444 m^2/s^2)
vf = 7.495 m/s

Therefore, the velocity with which he strikes the water is 7.495 m/s.

(b) What is the highest point he reaches above the water:

To find the highest point, we need to find the maximum height the diver reaches above the water, which occurs when his velocity becomes zero before he starts descending.

Using the equation of motion:

v^2 = u^2 + 2as

where
v = final velocity (0 m/s, becomes zero at the highest point)
u = initial velocity (1.66 m/s, positive since it's upward)
a = acceleration due to gravity (-9.8 m/s^2, negative since it's downward)
s = displacement (unknown)

Plugging in the given values, we can solve for the displacement:

0^2 = (1.66 m/s)^2 + 2*(-9.8 m/s^2)*s
0 = 2.7556 m^2/s^2 - 19.6 m/s^2 * s
19.6 m/s^2 * s = 2.7556 m^2/s^2
s = 2.7556 m^2/s^2 / 19.6 m/s^2
s = 0.14 m

Therefore, the highest point the diver reaches above the water is 0.14 m.