Animated Solution for Physics - Kinematics: A particle is projected at $60^{\circ }$ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is

Imagine standing on an open field, a ball in your hand. You throw it forward and upward. What happens? It doesn't just go straight up and come straight down, nor does it travel in a straight line. It traces a beautiful, symmetrical curve known as a parabola. This is the essence of projectile motion, a mesmerizing dance between inertia and gravity.

When a particle is projected at an angle, it possesses an initial velocity vector. This vector is the driving force of the motion, but it doesn't act alone. Gravity, the ever-present invisible hand, constantly pulls the particle downwards. However, gravity only cares about the vertical dimension. It has absolutely no influence on the horizontal motion of the particle. This independence of horizontal and vertical motion is the master key to unlocking projectile problems.

In our specific problem, the particle is launched at an angle of $60^{\circ }$ to the horizontal. It starts with a burst of energy, specifically a kinetic energy denoted as $K$. Our mission is to track this energy and find out what happens to it when the particle reaches the very peak of its flight.

To understand the energy, we first need to understand the velocity. The initial velocity, let's call it $v$, is launched at an angle $\theta =60^{\circ }$. Because velocity is a vector, we can break it down into two independent components: a horizontal component and a vertical component.

The horizontal component is given by $v_x=v\cos \theta$. This part of the velocity is responsible for moving the particle forward across the ground. The vertical component is $v_y=v\sin \theta$, and this is what drives the particle upwards against gravity.

As the particle ascends, gravity pulls it down, causing the vertical velocity $v_y$ to decrease steadily. It gets slower and slower in the upward direction until, for one fleeting moment at the highest point, it stops moving up entirely. At this exact peak, the vertical velocity is zero ($v_y=0$).

But here is the crucial catch: what about the horizontal velocity? Since there is no air resistance (unless stated otherwise) and gravity only acts vertically, there is absolutely no force pushing or pulling the particle horizontally. According to Newton's First Law, an object in motion stays in motion unless acted upon by a force. Therefore, the horizontal velocity remains perfectly constant throughout the entire flight!

At the highest point, the particle is still moving forward. Its total velocity is not zero; it is exactly equal to its horizontal component: $v_{\text{top}}=v\cos 60^{\circ }$.

Now, let's bring kinetic energy into the spotlight. Kinetic energy is the energy of motion, defined by the classic formula:

This is the energy the particle has at the very beginning of its journey. It depends on the mass $m$ and the square of the initial total velocity $v$.

Fast forward to the highest point. The particle is still moving, so it still has kinetic energy. Let's call this new kinetic energy $K_{\text{top}}$. We calculate it using the exact same formula, but we must use the velocity the particle has at that specific point.

We already established that $v_{\text{top}}=v\cos 60^{\circ }$. Let's substitute this into our kinetic energy equation:

When we square the terms inside the parenthesis, we get:

Look closely at the expression we just derived. Do you see a familiar friend hiding in there? The term $\frac{1}{2}m{v}^2$ is exactly our initial kinetic energy, $K$! We can substitute $K$ back into the equation to make it incredibly elegant:

This is a powerful realization. The kinetic energy at the highest point is simply the initial kinetic energy multiplied by the square of the cosine of the launch angle. This formula works for any projectile motion problem of this type!

Now, we just need to crunch the numbers. From our trigonometric tables, we know that the cosine of $60^{\circ }$ is $\frac{1}{2}$.

Substituting this value into our equation:

Squaring the fraction gives us:

And there we have it! The kinetic energy at the highest point is exactly one-quarter of the initial kinetic energy.

Energy Conservation

You might be wondering, if the kinetic energy dropped from $K$ to $\frac{K}{4}$, where did the missing $\frac{3K}{4}$ go? Did the universe just lose energy?

Absolutely not! The Law of Conservation of Energy dictates that energy cannot be created or destroyed, only transformed. As the particle climbed higher and higher, it was doing work against gravity. The kinetic energy it lost was seamlessly converted into gravitational potential energy.

At the highest point, the particle has reached its maximum altitude. Therefore, its potential energy is at its absolute maximum. The total mechanical energy of the system remains constant at $K$. So, at the peak:

This beautiful interplay between kinetic and potential energy is what makes physics so profoundly elegant. By understanding the components of velocity and the conservation of energy, you can conquer any projectile motion problem that comes your way!