Animated Solution for Physics - Kinematics: A grasshopper is sitting on the horizontal ground and the sun is shining at an angle $\varphi$ above the horizon. The grasshopper jumps towards the sun with an initial velocity $u$ at an angle $\theta$ with the ground. Find expression for speed of shadow of the grasshopper on the ground. Acceleration due to gravity is $g$.

A Leap Towards the Light

Imagine a peaceful meadow. A grasshopper sits on the ground, facing the sun which hangs at an angle $\varphi$ above the horizon. Suddenly, it leaps! It jumps towards the sun with an initial velocity $u$ at an angle $\theta$.

As the grasshopper traces a beautiful parabolic path through the air, the sun's rays cast its shadow on the ground. Our mission is to find the speed of this shadow. At first glance, you might think the shadow just moves along with the grasshopper. But as we'll see, the interplay between the grasshopper's projectile motion and the geometry of the sun's rays creates a fascinating, dynamic result.

To solve any kinematics problem, we must first establish a solid mathematical foundation. Let's set the grasshopper's starting point as the origin $(0,0)$ of our coordinate system.

As the grasshopper flies through the air, its position at any time $t$ is governed by the classic equations of projectile motion. The horizontal motion is uniform, unaffected by gravity:

The vertical motion, however, is a constant acceleration battle against gravity:

These two equations perfectly describe where the grasshopper is at any given moment. But we don't just want the grasshopper's position; we want the shadow's position.

Here is where the physics meets geometry. The sun is at an angle $\varphi$ above the horizon. Because the sun is incredibly far away, we can treat its rays as perfectly parallel lines striking the ground at this angle $\varphi$.

Consider a single ray of light that grazes the grasshopper at coordinates $(x,y)$ and hits the ground. This ray forms a right-angled triangle. The vertical height of this triangle is simply the grasshopper's height, $y$. The base of the triangle lies on the ground, stretching from the shadow's position $x_s$ to the point directly below the grasshopper, $x$.

Because the grasshopper jumped towards the sun, the sun is in front of it. The rays slant downwards and backwards, meaning the shadow falls behind the grasshopper. The length of the triangle's base is therefore $x-x_s$.

Using basic trigonometry on this right-angled triangle, we can relate the angle $\varphi$ to the sides:

Rearranging this to solve for the shadow's position $x_s$, we get:

This is our master equation! It elegantly links the shadow's position to the grasshopper's 2D coordinates and the sun's angle.

We have the position of the shadow, but we need its speed. In physics, whenever we need to go from position to speed, we call upon the power of calculus. We must differentiate the shadow's position $x_s$ with respect to time $t$.

Taking the time derivative of our master equation:

Notice how beautifully the chain rule applies here. The terms $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are simply the horizontal and vertical velocities of the grasshopper, $v_x$ and $v_y$. So, the shadow's velocity is:

Now, we substitute the known velocity components of the grasshopper. The horizontal velocity is constant:

The vertical velocity changes over time due to gravity:

Plugging these into our velocity equation gives:

We have the raw expression, but in physics, we always strive for elegance. Let's expand and simplify this equation. First, distribute the $\cot \varphi$:

To combine the first two terms, it helps to express the cotangent in terms of sine and cosine: $\cot \varphi =\frac{\cos \varphi }{\sin \varphi }$.

Now, find a common denominator for the terms inside the parenthesis:

Look closely at the numerator: $\sin \varphi \cos \theta -\cos \varphi \sin \theta$. This is a classic trigonometric identity! It perfectly collapses into $\sin (\varphi -\theta )$.

Substituting this back, we arrive at our final, beautiful expression for the shadow's speed:

This equation tells a profound story. The shadow's speed is not constant! Because of the $gt\cot \varphi$ term, the shadow actually accelerates across the ground. As the grasshopper falls faster and faster, the angled sun rays project that vertical acceleration into horizontal acceleration of the shadow.

Of course, this magical dance of light and motion only lasts as long as the grasshopper is airborne. The expression is valid only for the time of flight, $t<\frac{2u\sin \theta }{g}$. Once the grasshopper lands, the jump is over, and the shadow comes to rest.