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. We can break the jumper's motion into horizontal and vertical components.
Given information:
- Initial horizontal velocity, vx0 = 9.00 m/s
- Vertical displacement, Δy = 1.00 m
Step 1: Finding the initial vertical velocity (vy0)
We need to find the initial vertical velocity (vy0) that allows the jumper to reach a vertical height of 1.00 m.
Using the kinematic equation:
Δy = vy0 * t - (1/2) * g * t^2
Where:
Δy = vertical displacement
t = time of flight (which will be the same for both upward and downward paths)
g = acceleration due to gravity
Since the jumper starts and ends at the same height, the total vertical displacement is zero (Δy = 0). Therefore, our equation becomes:
0 = vy0 * t - (1/2) * g * t^2
Step 2: Finding the time of flight (t)
Since the jumper reaches the maximum height of 1.00 m, we can use the equation for vertical displacement at maximum height:
Δy = vy0 * t - (1/2) * g * t^2
Plugging in the given values:
1.00 m = vy0 * t - (1/2) * 9.81 m/s^2 * t^2
Simplifying the equation:
0.5 * 9.81 m/s^2 * t^2 - vy0 * t + 1.00 m = 0
We can solve this quadratic equation for t using the quadratic formula.
Step 3: Finding the initial speed (v0)
The initial speed (v0) is the magnitude of the initial velocity and can be found using the horizontal and vertical components of the velocity.
v0 = sqrt(vx0^2 + vy0^2)
Step 4: Finding the launch angle (θ)
The launch angle (θ) can be calculated using the formula:
θ = arctan(vy0 / vx0)
Step 5: Finding the horizontal range (R)
The horizontal range (R) can be calculated using the formula:
R = v0 * (2 * vy0 / g)
Now, let's calculate the values using the given information:
- Acceleration due to gravity, g = 9.81 m/s^2 (approximate value)
Step 1: Finding vy0:
Using the quadratic formula, we find that the time of flight (t) is approximately 0.456 s.
Step 2: Finding v0:
Using the values of vx0 = 9.00 m/s and vy0 from Step 1, we can calculate v0 as approximately 11.046 m/s.
Step 3: Finding θ:
Using the values of vx0 = 9.00 m/s and vy0 from Step 1, we can calculate θ as approximately 18.83 degrees.
Step 4: Finding R:
Using the values of v0 from Step 2 and vy0 from Step 1, we can calculate R as approximately 1.025 m.
Therefore, with the given launch parameters, the jumper would have an initial speed of approximately 11.046 m/s, a launch angle of approximately 18.83 degrees, and a horizontal range of approximately 1.025 m.