Juggling a ball I have the following exercise:

Explain the progress of the force you need to juggle a $5\ \rm{kg}$ ball as a function of time. 

Now since the question is kind of imprecise to me, I have problems solving it.
My approaches:
If I hold the ball in my hand, I need a force of $mg$ right? Because of gravity. Now if I want to throm it in the air I need to apply a force bigger than $mg$. Are these thoughts correct?
And now how to write this as a function of time?
 A: Since this is a homework-type question, I'm not going to write the entire answer:
In juggling, the position of balls with respect to time looks like this.
Let's denote the length of your forearm $d$, the breadth of your shoulders $b$, let's assume the highest point the balls achieve is at your chin, and let's denote the height difference between your chin and your hand at the moment of release $h_r$, at the moment of catch $h_c$, the height difference between your chin and your elbow $h$ and the angle between your forearm and a horizontal line $\theta$ (we'll assume $\theta$ is small).
$$h_r = h + d\sin\theta\\
h_c = h - d\sin\theta$$
Let the time the ball is flying upwards $t_1$, the time downwards $t_2$ and the net time $t$:
$$t_1 = \sqrt\frac{2(h+d\sin\theta)}{g}\\
t_2 = \sqrt\frac{2(h-d\sin\theta)}{g}\\
t = \sqrt\frac2g(\sqrt{h+d\sin\theta} + \sqrt{h - d\sin\theta})$$
Let's denote the horizontal velocity during the travel through the air as $v_x$, vertical velocity when leaving one hand $v_{y1}$ and when being caught by the other hand $v_{y2}$:
$$v_x = \frac{b}t = \frac{b\sqrt\frac{g}2}{(\sqrt{h+d\sin\theta} + \sqrt{h - d\sin\theta})}\\
v_{y1}=gt_1 = \sqrt{2g(h+d\sin\theta)}\\
v_{y2}=gt_2 = \sqrt{2g(h-d\sin\theta)}$$
The speed $v_c$ when being caught is (we approximate $\sin^2\theta\approx0$)
$$v_c = \sqrt{v_x^2+v_{y2}^2}=\sqrt{\frac{b^2g+16gh(h-\sin\theta)}{8h}}$$
and the speed $v_r$ when released (again we approximate $\sin^2\theta\approx0$)
$$v_r = \sqrt{v_x^2+v_{y1}^2}=\sqrt{\frac{b^2g+16gh(h+\sin\theta)}{8h}}$$
Let's denote $T$ the period of the motion and assume that during the time the ball is in your hand, its speed increases linearly from $v_c$ to $v_r$:
$$\pi d = v_c\frac T 2 + \frac12\frac{v_r-v_c}{\frac T 2}\left(\frac T 2\right)^2$$
From that we have
$$T = \frac{4\pi d}{v_c + v_r}$$
Assume that during the time the ball is in your hand, its speed increases linearly from $v_c$ to $v_r$, the dependence of the speed on time is therefore:
$$v(t) = v_c + \frac{v_r - v_c}{\frac T 2}t = v_c + \frac{v_r^2 - v_c^2}{2\pi d}t$$
For your next steps, look up centripetal force.
