Animated Solution for Physics - Kinematics: A small ball is thrown from foot of a wall with the minimum possible velocity to hit a bulb B on the ground a distance $L$ away from the wall. Find expression for height $h$ of shadow of the ball on the wall as a function of time $t$. Acceleration due to gravity is $g$.

Physics is often full of beautiful surprises where complex, accelerating motions perfectly align to create something incredibly simple. This problem is a classic example of such mathematical elegance, blending the kinematics of a projectile with the coordinate geometry of straight lines.

Imagine you are standing at the foot of a wall (the origin, $x=0,y=0$) and you need to throw a ball to hit a bulb resting on the ground at a distance $L$ away (coordinates $L,0$). The problem imposes a strict condition: you must use the minimum possible velocity.

From the standard projectile range formula, $R=\frac{u^2\sin (2\theta )}{g}$, we know that to achieve a given range $R=L$ with the smallest possible initial speed $u$, we must maximize the sine function. This happens when $\sin (2\theta )=1$, which means the angle of projection must be exactly $\theta =45^{\circ }$.

Plugging this back in, we find the initial velocity squared is $u^2=gL$, giving $u=\sqrt{gL}$. Because the angle is $45^{\circ }$, the horizontal and vertical components of this velocity are identical:

With the initial velocities locked in, we can write down the exact coordinates of the ball $(x,y)$ at any given time $t$ using the equations of motion:

Now comes the crucial geometric insight. How is the shadow formed on the wall? A light ray originates from the bulb at $(L,0)$, travels in a perfectly straight line, passes exactly through the ball at $(x,y)$, and strikes the wall at a height $(0,h)$.

Because light travels in straight lines, these three points—the bulb, the ball, and the shadow—must be collinear. This means the slope of the line segment connecting the shadow to the bulb is identical to the slope of the line segment connecting the ball to the bulb.

Equating the slopes:

Rearranging this gives us a direct expression for the shadow's height:

We now substitute our time-dependent expressions for $x(t)$ and $y(t)$ into this geometric relation:

At first glance, this looks like a messy, non-linear function of time. But let's factor it carefully. In the numerator, we can pull out $t\sqrt{\frac{g}{2}}$:

In the denominator, we can factor out $\sqrt{L}$:

Notice the magic? The entire complex bracket $\left(\sqrt{L}-t\sqrt{\frac{g}{2}}\right)$ appears in both the numerator and the denominator. It cancels out perfectly!

After the cancellation, we are left with a stunningly simple result:

The height of the shadow is directly proportional to time $t$. This means that while the ball is undergoing a complex parabolic trajectory, constantly accelerating downwards due to gravity, its shadow is marching up the wall with a perfectly constant velocity of $v_h=\sqrt{\frac{gL}{2}}$. It is a beautiful reminder of how underlying symmetries in physics can lead to incredibly elegant outcomes.