The object of the long jump is to launch oneself as a

projectile and attain the maximum horizontal range
(Fig. 3–26). Here we shall treat the long jumper as a
particle even though the human body is fairly large
compared to the size of the trajectory. Actually there is
one point within the athlete’s body, called the “center of
mass” (to be studied in Chapter 5), that behaves as a
projected particle. Our analysis of projectile motion
implies that the long jumper should try to maximize v0
and take off at an angle as close to 45.0� as possible.
However, it is easier to get a large value of vx0 (by a
running start) than it is to get a large value of vy 0; consequently
'0 is usually much less than 45.0�. Suppose the
jumper takes off with vx0
9.00 m/s and jumps with a
value of vy0 sufficient to reach a vertical height of 1.00 m.
Find v0, '0, and the horizontal range. The world record,
as of 1994, is 8.95 m.

To solve this problem, we need to use the principles of projectile motion. Here is how we can find the answers to the given question:

1. Calculating the initial velocity (v0):
We are given the horizontal component of the initial velocity (vx0 = 9.00 m/s). To find the magnitude of the initial velocity (v0), we can use the following equation:

v0 = sqrt((vx0)^2 + (vy0)^2)

We are not given the value of the vertical component of the initial velocity (vy0), but we know that the jumper jumps with a value of vy0 sufficient to reach a vertical height of 1.00 m. At the highest point of the jump, the vertical component of the velocity becomes zero. Using the equation for vertical motion:

vy0 = sqrt(2 * g * h)

where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the vertical height (1.00 m). Plugging in these values, we can find vy0.

2. Calculating the launch angle ('0):
The launch angle ('0) is the angle at which the jumper takes off. Since '0 is the angle between the initial velocity vector and the horizontal, we can find it using the equation:

'0 = atan(vy0 / vx0)

where atan is the inverse tangent function.

3. Calculating the horizontal range:
The horizontal range is the total distance covered by the jumper in the horizontal direction. We can find it using the equation:

R = (v0^2 * sin(2 * '0)) / g

where R is the horizontal range.

After finding v0, '0, and R, we can compare the horizontal range (R) to the world record (8.95 m) to see if it has been exceeded.

Now, let's plug in the given values and solve the problem:

1. Calculating v0:
vy0 = sqrt(2 * 9.8 * 1) = 4.43 m/s
v0 = sqrt((9.00)^2 + (4.43)^2) = 10.13 m/s

2. Calculating '0:
'0 = atan(4.43 / 9.00) = 25.11 degrees

3. Calculating the horizontal range:
R = (10.13^2 * sin(2 * 25.11)) / 9.8 = 8.94 m

Comparing the calculated horizontal range (8.94 m) to the world record (8.95 m), we see that it is very close but slightly less than the world record.