The motion of a human body through space can be modeled as the motion of a particle at the body's center of mass. The components of the position of an athlete's center of mass from the beginning to the end of a certain jump are described by the following two equations, where t is the time at which the athlete lands after taking off at t = 0.

xf = 0 + (9.6 m/s)(cos 18.5°)t
0.140 m = 0.710 m + 0 + (9.6 m/s)(sin 18.5°)t - (9.80 m/s2)t2
(a) Identify his vector position.
( i + j) m
(b) Identify his vector velocity at the takeoff point.
m/s °
(c) The world longjump record is 8.95 m. How far did the athlete jump in this problem?
(d) Describe the shape of the trajectory of his center of mass.

no answer

(a) The vector position of the athlete's center of mass can be expressed as:

r = xf i + yf j

Where xf and yf are the components of the position in the x and y direction, respectively.

(b) The vector velocity of the athlete at takeoff can be found by taking the derivative of the position vector with respect to time:

v = dx/dt i + dy/dt j

Given the equations:

xf = (9.6 m/s)(cos 18.5°)t
0.140 m = 0.710 m + (9.6 m/s)(sin 18.5°)t - (9.80 m/s^2)t^2

Differentiating xf with respect to t:

dx/dt = (9.6 m/s)(cos 18.5°)

Differentiating the second equation with respect to t:

dy/dt = (9.6 m/s)(sin 18.5°) - 2(9.80 m/s^2)t

Therefore, the vector velocity at takeoff is:

v = (9.6 m/s)(cos 18.5°) i + [(9.6 m/s)(sin 18.5°) - 2(9.80 m/s^2)(t_takeoff)] j

(c) To determine how far the athlete jumped, we can find the value of xf at the time when the athlete lands. Set yf = 0. Note that the athlete lands when t = t_landing. Therefore, the equation becomes:

0 = 0.71 m + (9.6 m/s)(sin 18.5°)(t_landing) - (9.80 m/s^2)(t_landing)^2

Solve this equation to find the value of t_landing. Once t_landing is known, substitute it into the equation for xf to find the distance traveled in the x-direction.

(d) The shape of the trajectory of the athlete's center of mass can be determined from the equation. Since the motion is described by two separate equations for xf and yf, we can consider it a parabolic trajectory in two dimensions.

(a) To identify the vector position, we need to find the values of the x and y components of the position.

From the given equation, xf = 0 + (9.6 m/s)(cos 18.5°)t, we can see that the x-component of the position is given by 0 + (9.6 m/s)(cos 18.5°)t.

From the second equation, 0.140 m = 0.710 m + 0 + (9.6 m/s)(sin 18.5°)t - (9.80 m/s^2)t^2, we can see that the y-component of the position is given by 0.140 m - 0.710 m - (9.6 m/s)(sin 18.5°)t + (9.80 m/s^2)t^2.

So, the vector position is given by (0 + (9.6 m/s)(cos 18.5°)t) i + (0.140 m - 0.710 m - (9.6 m/s)(sin 18.5°)t + (9.80 m/s^2)t^2) j.

(b) To identify the vector velocity at the takeoff point, we need to find the values of the x and y components of the velocity.

The x-component of the velocity can be found by taking the derivative of the x-component of the position with respect to time (t).

The y-component of the velocity can be found by taking the derivative of the y-component of the position with respect to time (t).

(c) To find how far the athlete jumped, we need to find the value of xf when the athlete lands. We can set the y-component of the position equation equal to zero and solve for t. Then, we substitute this value of t into the x-component of the position equation to find xf.

(d) To describe the shape of the trajectory of the athlete's center of mass, we can analyze the equation for the y-component of the position. The presence of t^2 term suggests a parabolic trajectory.

. The motion of a human body through space can be

modeled as the motion of a particle at the body’s cen-
ter of mass as we will study in Chapter 9. The compo-
nents of the displacement of an athlete’s center of mass
from the beginning to the end of a certain jump are
described by the equations
xf 5 0 1 (11.2 m/s)(cos 18.5°)t
0.360 m 5 0.840 m 1 111.2 m/s2 1sin 18.582t 2 1
2 19.80 m/s
2 2t
2
where t is in seconds and is the time at which the ath-
lete ends the jump. Identify (a) the athlete’s position
and (b) his vector velocity at the takeoff point. (c) How
far did he jump?