Animated Solution for Physics - Kinematics: A ball is thrown vertically upwards. Its distance $s$ from a fixed point varies with time $t$ according to the following graph. Calculate velocity of projection of the ball.

Imagine you are standing on the ground, watching a ball being thrown straight up into the air. If you were to measure the distance from the ball to your eye, that distance would constantly change. But what if you plotted that distance against time, and the graph looked like a roller coaster with two distinct dips? That is exactly the puzzle we are facing in this beautiful kinematics problem.

The first thing that should catch your eye when looking at the given $s-t$ graph is that the distance $s$ never drops to zero. The absolute minimum distance recorded is $s_3=6\text{ m}$.

Think about what this physically implies. If the fixed point were located anywhere on the vertical line along which the ball is traveling, the ball would eventually pass right through it. At that exact instant, the distance would be zero. Because the distance never becomes zero, we can definitively conclude that the fixed point is horizontally offset from the line of motion.

Let's set up a coordinate system to make sense of this. Assume the ball is thrown from the origin $(0,0)$ and moves purely along the $y$-axis. Its position at any time $t$ is simply $(0,y)$.

Now, let the mysterious fixed point be located at coordinates $(x_0,y_0)$. Using the standard Euclidean distance formula, the distance $s$ between the ball and the fixed point at any instant is given by:

This single equation is the master key to unlocking the entire problem. We just need to evaluate it at the critical points shown on the graph.

Finding the Horizontal Offset

Look at the two dips in the graph where the distance hits its minimum value of $s_3=6\text{ m}$. Mathematically, looking at our master equation, $s(t)$ is minimized when the squared term $(y(t)-y_0{)}^2$ is as small as possible. Since a square cannot be negative, its minimum value is zero, which occurs exactly when $y(t)=y_0$.

Physically, this means the ball is exactly at the same height as the fixed point. At this instant, the vertical separation vanishes, and the distance $s$ is purely the horizontal offset:

So, we now know the fixed point is $6\text{ m}$ away horizontally!

Reaching the Peak

Between the two dips, the graph rises to a local maximum of $s_2=10\text{ m}$. Why does it go up and then back down? Because the ball continues to travel upward past the height $y_0$, reaches its maximum height $H$, and then falls back down, passing $y_0$ a second time!

At the very peak of its trajectory, the ball's height is $y=H$. Plugging this into our master equation gives:

We already know $x_0=6$. Substituting this in, we get:

This forms a classic $6-8-10$ Pythagorean triplet! Solving for the vertical separation between the peak and the fixed point, we find:

Finding the Vertical Offset

We are almost there. We know the peak is $8\text{ m}$ above the fixed point, but how high is the fixed point itself? To find $y_0$, we look at the very beginning of the motion.

At $t=0$, the ball is launched from the ground, so its height is $y=0$. The graph tells us that the initial distance to the fixed point is $s_1=6\sqrt{5}\text{ m}$. Let's plug these initial conditions into our master equation:

Substituting $x_0=6$ and squaring both sides:

The fixed point is located $12\text{ m}$ above the ground!

Now we can piece everything together. The fixed point is at a height of $12\text{ m}$, and the ball travels an additional $8\text{ m}$ above that to reach its peak. Therefore, the total maximum height $H$ reached by the ball is:

Finally, we return to pure 1D kinematics. The maximum height $H$ of a projectile thrown vertically upwards with initial velocity $u$ is given by the standard formula:

Plugging in our known values ($H=20\text{ m}$ and $g=10{\text{ m/s}}^2$):

The velocity of projection of the ball is $20\text{ m/s}$. By carefully translating a 1D motion into a 2D geometric distance formula, we decoded the graph and uncovered the hidden kinematics. A truly elegant problem!