Solution 15
Step 1: Kinematic Setup
The condition “minimum possible velocity” to reach horizontal distance $L$ implies a projection angle of $\theta = 45^\circ$. $$ L = \frac{u^2 \sin(2\theta)}{g} = \frac{u^2}{g} \implies u^2 = gL \quad \text{and} \quad u = \sqrt{gL} $$ The velocity components are: $$ u_x = u \cos 45^\circ = \sqrt{gL} \cdot \frac{1}{\sqrt{2}} $$ $$ u_y = u \sin 45^\circ = \sqrt{gL} \cdot \frac{1}{\sqrt{2}} $$ The position coordinates at time $t$ are: $$ x = u_x t = t\sqrt{\frac{gL}{2}} $$ $$ y = u_y t – \frac{1}{2}gt^2 = t\sqrt{\frac{gL}{2}} – \frac{1}{2}gt^2 $$
The condition “minimum possible velocity” to reach horizontal distance $L$ implies a projection angle of $\theta = 45^\circ$. $$ L = \frac{u^2 \sin(2\theta)}{g} = \frac{u^2}{g} \implies u^2 = gL \quad \text{and} \quad u = \sqrt{gL} $$ The velocity components are: $$ u_x = u \cos 45^\circ = \sqrt{gL} \cdot \frac{1}{\sqrt{2}} $$ $$ u_y = u \sin 45^\circ = \sqrt{gL} \cdot \frac{1}{\sqrt{2}} $$ The position coordinates at time $t$ are: $$ x = u_x t = t\sqrt{\frac{gL}{2}} $$ $$ y = u_y t – \frac{1}{2}gt^2 = t\sqrt{\frac{gL}{2}} – \frac{1}{2}gt^2 $$
Step 2: Shadow Geometry
The light source is at $x=L, y=0$. The wall is at $x=0$. The ball is at $(x,y)$. Using similar triangles formed by the light ray: $$ \frac{h}{L} = \frac{y}{L-x} $$ $$ h = \frac{L y}{L – x} $$
The light source is at $x=L, y=0$. The wall is at $x=0$. The ball is at $(x,y)$. Using similar triangles formed by the light ray: $$ \frac{h}{L} = \frac{y}{L-x} $$ $$ h = \frac{L y}{L – x} $$
Step 3: Substitution and Simplification
Substitute $x$ and $y$ into the expression for $h$: $$ h(t) = \frac{L \left( t\sqrt{\frac{gL}{2}} – \frac{1}{2}gt^2 \right)}{L – t\sqrt{\frac{gL}{2}}} $$ To clean this up, multiply the numerator and denominator by 2 to clear the fraction in the $y$ term: $$ h(t) = \frac{L (2t\sqrt{\frac{gL}{2}} – gt^2)}{2L – 2t\sqrt{\frac{gL}{2}}} $$ We can express $\sqrt{\frac{gL}{2}}$ as $\frac{\sqrt{2gL}}{2}$. Let’s simplify differently. Factor out $u_x = \sqrt{gL/2}$: $$ h(t) = \frac{L \left[ t\sqrt{\frac{gL}{2}} – \frac{gt^2}{2} \right]}{L – t\sqrt{\frac{gL}{2}}} $$ Recall $L = \frac{u^2}{g} = \frac{gL}{g} = L$. This is consistent. Let’s factor $t$ from the numerator: $$ h(t) = \frac{Lt \left( \sqrt{\frac{gL}{2}} – \frac{gt}{2} \right)}{L – t\sqrt{\frac{gL}{2}}} $$ Multiplying numerator and denominator by $\sqrt{2}$ to remove the root in the denominator term usually makes it cleaner: $$ h(t) = \frac{Lt (\sqrt{gL} – \frac{gt}{\sqrt{2}})}{L\sqrt{2} – t\sqrt{gL}} $$ This is a valid final expression. However, leaving it in terms of the initial constants is often standard.
Substitute $x$ and $y$ into the expression for $h$: $$ h(t) = \frac{L \left( t\sqrt{\frac{gL}{2}} – \frac{1}{2}gt^2 \right)}{L – t\sqrt{\frac{gL}{2}}} $$ To clean this up, multiply the numerator and denominator by 2 to clear the fraction in the $y$ term: $$ h(t) = \frac{L (2t\sqrt{\frac{gL}{2}} – gt^2)}{2L – 2t\sqrt{\frac{gL}{2}}} $$ We can express $\sqrt{\frac{gL}{2}}$ as $\frac{\sqrt{2gL}}{2}$. Let’s simplify differently. Factor out $u_x = \sqrt{gL/2}$: $$ h(t) = \frac{L \left[ t\sqrt{\frac{gL}{2}} – \frac{gt^2}{2} \right]}{L – t\sqrt{\frac{gL}{2}}} $$ Recall $L = \frac{u^2}{g} = \frac{gL}{g} = L$. This is consistent. Let’s factor $t$ from the numerator: $$ h(t) = \frac{Lt \left( \sqrt{\frac{gL}{2}} – \frac{gt}{2} \right)}{L – t\sqrt{\frac{gL}{2}}} $$ Multiplying numerator and denominator by $\sqrt{2}$ to remove the root in the denominator term usually makes it cleaner: $$ h(t) = \frac{Lt (\sqrt{gL} – \frac{gt}{\sqrt{2}})}{L\sqrt{2} – t\sqrt{gL}} $$ This is a valid final expression. However, leaving it in terms of the initial constants is often standard.
Final Answer:
$$ h(t) = \frac{Lt \sqrt{gL/2} – \frac{1}{2}gLt^2}{L – t\sqrt{gL/2}} $$ or simplifying the square roots: $$ h(t) = t \sqrt{gL/2} $$
$$ h(t) = \frac{Lt \sqrt{gL/2} – \frac{1}{2}gLt^2}{L – t\sqrt{gL/2}} $$ or simplifying the square roots: $$ h(t) = t \sqrt{gL/2} $$