Question

The motion of a human body through space can be modeled as the motion of a particle at the body's center of mass as we will study in a later chapter. The components 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

x_f = 0 + (10.8 m/s)(cos 18.5

Answer 1

For an "athlete jump", we are talking about the ballistic trajectory of his center of mass.

y=y0+vy0*t-g*t^2/2,
where vy0 is initial vertical velocity, y is the vertical position, and y0 is the initial vertical position.
Of course,
vy0=v0*sin(theta),
where theta is angle between initial velocity and horizontal (positive x-axis), while v0 is initial speed.
This should be familiar to you, and comes from the fact that:
vy=vy0-g*t
(gravity points downwards, and vy0 is measured as component vertically upwards, hence the minus sign between them).

On the other hand,
x=x0+vx0*t.
where
vx0=v*cos(theta) is the initial horizontal velocity. It does not change in time, as there is no horizontal force. Also, vx0 is measured as component of initial velocity along positive x-axis.

Part (a): Initial time is t=0, in the above equations. Your answer is correct.
Part (b): Initial velocity is vector
v0=(vx0,vy0),
Magnitude (speed) is sqrt(vx0^2+vy0^2), should be 10.8m/s, as your problem already gives the form with
vy0=v0*sin(theta)
vx0=v0*cos(theta),
.
Part (c): You need the horizontal distance that the athlete traveled, betwee the initial jump, and his impact with the ground at the final time. Obviously his center of mass is at height
y0=0.78m.
So, I guess it is correct to assume he lands when his center of mass again reaches the same height, at the end of the jump (he lands on his feet standing up, not on his bottom, as athletes actually do).
So you need the time at which
y(tf)=yf=0.78m=y0,
where tf is the final time (time of end of jump).
You can change this choice, maybe say
y(tf)=0, depending on how your problem wants you to model the jump.

Find tf from this equation (solve quadratic equation for value tf>0, because t=0 is initial time. Once you have it, plug it into equation for x, to get the horizontal distance traveled (the length of the jump), x(tf).