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 find the initial velocity (v0), launch angle ('0), and the horizontal range of the long jumper, we can use the known values of vx0, vy0, and the vertical height (h) reached.

Step 1: Resolve the initial velocity into its horizontal and vertical components.
- The horizontal component is vx0 = 9.00 m/s (given).
- The vertical component is vy0, which we need to calculate.

Step 2: Use the vertical motion equations to find vy0.
- We know that the jumper reaches a vertical height of h = 1.00 m.
- The equation for the vertical displacement (Δy) is given by:
Δy = vy0 * t + (1/2) * g * t^2 (where g is the acceleration due to gravity)

- At the maximum height, the vertical velocity (vy) is zero.
vy = vy0 - g * t

- To find the time of flight (t), we can use the vertical motion equation for velocity:
vy = vy0 - g * t

- Setting vy = 0 and solving for t:
0 = vy0 - g * t
t = vy0 / g (equation 1)

- Substituting the value of t into the equation for vertical displacement:
h = vy0 * (vy0 / g) + (1/2) * g * (vy0 / g)^2

- Simplifying the equation:
h = vy0^2 / g - (1/2) * vy0^2 / g
2h = vy0^2 / g
vy0^2 = 2gh
vy0 = √(2gh) (equation 2)

Step 3: Calculate v0 and '0.
- The total initial velocity (v0) can be found using the horizontal and vertical components:
v0 = √(vx0^2 + vy0^2)

- The launch angle ('0) can be determined using the inverse tangent function:
'0 = tan^(-1)(vy0 / vx0)

Step 4: Find the horizontal range (R).
- The horizontal range can be calculated using the time of flight and the horizontal velocity:
R = vx0 * t

Now let's use these steps to solve for v0, '0, and the horizontal range.

Given:
vx0 = 9.00 m/s
h = 1.00 m
g = 9.8 m/s^2 (approximation for acceleration due to gravity)

Step 1: vx0 is already given as 9.00 m/s.

Step 2: Calculate vy0.
Using equation 2:
vy0 = √(2 * 9.8 * 1)
vy0 = √(19.6)
vy0 ≈ 4.43 m/s

Step 3: Calculate v0 and '0.
Using the formula for v0:
v0 = √(9.00^2 + 4.43^2)
v0 ≈ 9.97 m/s

Using the inverse tangent function for '0:
'0 = tan^(-1)(4.43 / 9.00)
'0 ≈ 25.67 degrees

Step 4: Calculate the horizontal range (R).
Using equation 1 to find the time of flight:
t = vy0 / g
t = 4.43 / 9.8
t ≈ 0.452 seconds

Using the formula for the horizontal range:
R = 9.00 * 0.452
R ≈ 4.08 m

The long jumper's initial velocity (v0) is approximately 9.97 m/s, the launch angle ('0) is approximately 25.67 degrees, and the horizontal range (R) is approximately 4.08 m.