Distance-time graph of a ball throw vertically up from a fixed point

[brotherbobby]

Homework Statement

A ball is thrown vertically up. Its distance from a fixed point varies with time according to the given graph (see below). Calculate the $\text{velocity of projection}$ of the ball.

Relevant Equations

For uniformly accelerated motion under gravity, taking $g = +9.8\;\text{m/s}^2$, we have the position after a time $t$, $y(t) = y_0+v_0t-\frac{1}{2}gt^2\;\text{(I)}$, and the velocity $v(t)=v_0-gt\; \text{(II)}$. The velocity can also be expressed as a funtion of the position $y$ from the origin, $v^2(y)=v^2_0-2gy\;\text{(III)}$.

Additionally, if the given fixed point doesn't lie along the path of motion of the ball, then $\vec s=\vec{r}-\vec {r'}$, where $\vec s$ is the position vector of the ball from the point and $\vec{r},\vec{r'}$ are those of the ball and the point from some origin (say from the point of projection).

Statement of the problem : I copy and paste the statement of the problem to the right as it appeared on the website. Given below is the graph of the ball as its distance from a fixed point with time.

Attempt : Where does this fixed point, say $\text{P}$ lie?

Imagine the fixed point lied along the path of motion of the ball. It can't be below the point of projection $\text{O}$, as distance of separation decreases initially. It can't be somewhere between $\text{O}$ and $\text{H}$, the maximum height, since in that case the distance should go to 0 for some time $t$ where the ball passes through the point. It can lie above the maximum height $\text{H}$. In that case, the distance of separation will reduce for some time and begin to increase again, as it does on the graph. But it should go up to the same value for some time $t$ later as for $t=0$ which it does not!

I suspect the fixed point of observation lies outside the path of motion of the ball, as shown in the diagram below. Additionally, for reference, I label points along the graph.

I put points along the motion of the ball that correspond to those in the graph in yellow on black background.

But this diagram cannot be correct.

From the graph, at B, where the distance from P is minimum, the body is at rest (momentarily), meaning its distance of separation isn't changing with time. But from my motion diagram on the right, the distance of separation $s$ is changing with time. So they contradict. Likewise for D.
If you put the observation point P anywhere else, it would violate the distance of separation as given in the graph.

I couldn't do better than this.

A hint or suggestion would be very welcome.

 

[Orodruin]

That the change in distance to a fixed point momentarily has zero derivative does not mean that the object is at rest. Consider uniform circular motion for example: the distance to the center is constant but the object undergoing the circular motion is certainly not at rest.

 

[brotherbobby]

For (uniform) circular motion, the velocity vector $\vec v_0$ is everywhere perpendicular to the radius vector $\vec{r}$. This vector is also the distance of separation $s$. Because of this perpendicularity, $\dot s=0$ though $v=v_0\ne 0$. Note the motion doesn't have to be uniform for this to be true, but circular.
For this problem, I should draw things more clearly showing the vector separation $\vec{s}$.

For the point B, the separation vector $\vec s$ is $\perp ^{r}$ to motion $\vec r'(\mathbf B$) and therefore $\dot s(\mathbf B)=0$. At C, the separation vector is clearly not perpendicular to motion. Is $\dot s(\mathbf C)=0$? Yes. This is because $\vec s=\vec r'-\vec r$. Since $\dot r=0$ always and $\dot r'(\mathbf C) = 0$, we have $\dot s(\mathbf C)=0$. Thus P is a correct point for the motion.

How do I solve the problem now? I have to find the velocity of projection $v_0$.

Let us have the graph.

At the point A (time of launch), we have $\vec s(\mathbf A)=\vec r$. But it's given $s(\mathbf A)=s_1=6\sqrt 5= r$. At B, $s(\mathbf B)=s_3=6$. From the right angled triangle in the diagram above on the right, it must mean that at for the point B, $r'(\mathbf B) = \sqrt{r^2-s^2(\mathbf B)}=\sqrt{36\times 5-36}= 12$. Likewise for the highest point C. I draw the motion diagram below.

For the right angled triangle PBC, $\mathbf{BC} = \sqrt{10^2-6^2}=8$.
Hence the maximum height attained by the ball $\mathbf{AC} = 20=H$.
Using $v_0^2=2gH\Rightarrow v_0=\sqrt{2\times 10\times 20} = \boxed{20}\quad\large{\color{green}\checkmark}$

 

[Delta2]

Can someone give me a definition of what exactly do we mean by velocity of projection of the ball, from what I understand we calculate the maximum velocity and we call it projection velocity for some reason?

Also is this problem all about use some sort of what I call ad hoc logic or should I say heuristic logic to determine where exactly in the plane the point P is, and where the rest of the points are along the vertical trajectory of the ball.

 

[brotherbobby]

Can someone give me a definition of what exactly do we mean by velocity of projection of the ball, from what I understand we calculate the maximum velocity and we call it projection velocity for some reason?

The speed with which the ball is projected. Same as the initial speed, which I write as $v_0$. It need not be the maximum speed though. Think about a ball projected vertically up from the top of a tower with speed $v_0$. By the time it hits the ground, its speed $v_G>v_0$.

Also is this problem all about use some sort of what I call ad hoc logic or should I say heuristic logic to determine where exactly in the plane the point P is, and where the rest of the points are along the vertical trajectory of the ball.

Yes, in this problem, P lies in the same plane in which the motion is taking place. If it did not, it could be a much harder problem.

 

[Delta2]

The speed with which the ball is projected. Same as the initial speed,

Oh I see because projection in mathematics have different meaning, I thought somehow it was meant that we take the projection of the ball to some other plane or line.