From a point on the ground at a distance of x from the foot of a vertical wall, a ball is thrown at an angle of 45° which just clears the top of the wall and after wards strike the ground at a…

Solution to the Projectile Motion Problem

Given:

• A ball is thrown from a point on the ground at a distance $x$ from the foot of a vertical wall.
• The angle of projection is $\theta =45^{\circ }$.
• The ball just clears the top of the wall.
• The ball strikes the ground at a distance $y$ on the other side of the wall.
• Let $h$ be the height of the wall.

To Prove: $$h=\frac{xy}{x+y}$$

Step 1: Establish the Equation of the Trajectory

Let the point of projection be the origin $O(0,0)$. The horizontal axis is the ground, and the vertical axis is the wall's vertical line (though the wall is at distance $x$).

The equation of the path of a projectile projected with initial velocity $u$ at an angle $\theta$ is given by: $$y_{coord}=x_{coord}\tan \theta -\frac{g{x}_{coord}^2}{2u^2{\cos }^2\theta }$$

Here, to avoid confusion with the given distances $x$ and $y$, let us use standard coordinates $(X,Y)$ for the trajectory equation. $$Y=X\tan \theta -\frac{g{X}^2}{2u^2{\cos }^2\theta }$$

Given $\theta =45^{\circ }$:

• $\tan 45^{\circ }=1$
• $\cos 45^{\circ }=\frac{1}{\sqrt{2}}⟹{\cos }^{2}45^{\circ }=\frac{1}{2}$

Substituting these values into the trajectory equation: $$Y=X(1)-\frac{g{X}^2}{2u^2(1/2)}$$ $$Y=X-\frac{g{X}^2}{2u^2}\dots \text{(Equation 1)}$$

Let $k=\frac{g}{2u^2}$. Then, Equation 1 becomes: $$Y=X-k{X}^2\dots \text{(Equation 2)}$$

Step 2: Apply Conditions for the Wall and the Landing Point

Condition 1: The ball clears the top of the wall. The wall is located at a horizontal distance $x$ from the point of projection. The height of the wall is $h$. Since the ball just clears the top, the point $(x,h)$ lies on the trajectory.

Substitute $X=x$ and $Y=h$ into Equation 2: $$h=x-k{x}^2\dots \text{(Equation 3)}$$

Condition 2: The ball strikes the ground on the other side. The ball strikes the ground at a distance $y$ on the other side of the wall. The total horizontal distance from the point of projection to the landing point is the sum of the distance to the wall and the distance from the wall to the landing point. Total horizontal range $R=x+y$.

At the landing point, the height $Y=0$. The horizontal coordinate is $X=x+y$.

Substitute $X=x+y$ and $Y=0$ into Equation 2: $$0=(x+y)-k(x+y{)}^2$$ $$k(x+y{)}^2=x+y$$

Since $x$ and $y$ are non-zero distances, we can divide by $(x+y)$: $$k(x+y)=1$$ $$k=\frac{1}{x+y}\dots \text{(Equation 4)}$$

Step 3: Substitute and Solve for Height $h$

Now, substitute the value of $k$ from Equation 4 into Equation 3.

From Equation 3: $$h=x-k{x}^2$$

Substitute $k=\frac{1}{x+y}$: $$h=x-\left(\frac{1}{x+y}\right)x^2$$

Simplify the expression: $$h=x-\frac{x^2}{x+y}$$

Take $x$ as a common factor: $$h=x\left(1-\frac{x}{x+y}\right)$$

Combine the terms inside the bracket: $$h=x\left(\frac{x+y-x}{x+y}\right)$$ $$h=x\left(\frac{y}{x+y}\right)$$

$$h=\frac{xy}{x+y}$$

Conclusion: Thus, it is proved that the height of the wall is $\frac{xy}{x+y}$.

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Alternative Solution: Conservation of Mechanical Energy

Definitions:

1. Kinetic Energy (KE): The energy possessed by a body due to its motion. For a body of mass $m$ moving with velocity $v$, $KE=\frac{1}{2}m{v}^2$.
2. Potential Energy (PE): The energy possessed by a body due to its position or configuration. For a body of mass $m$ at a height $h$ above a reference level, $PE=mgh$.

Proof: Consider a body of mass $m$ falling freely from a height $H$ above the ground. Let the acceleration due to gravity be $g$. We analyze the total mechanical energy (Sum of KE and PE) at three positions: A (top), B (intermediate), and C (ground).

1. At Position A (Top, Height $H$):

• The body is just starting to fall, so initial velocity $v_A=0$.
• $K{E}_A=\frac{1}{2}m(0{)}^2=0$
• $P{E}_A=mgH$
• Total Energy $E_A=K{E}_A+P{E}_A=0+mgH=mgH$

2. At Position B (Intermediate, Height $h$):

• Let the body have fallen a distance $H-h$.
• Let the velocity at this point be $v_B$.
• Using the equation of motion $v^2=u^2+2as$: $$v_B^2=0^2+2g(H-h)=2g(H-h)$$
• $K{E}_B=\frac{1}{2}m{v}_B^2=\frac{1}{2}m[2g(H-h)]=mg(H-h)$
• $P{E}_B=mgh$
• Total Energy $E_B=K{E}_B+P{E}_B=mg(H-h)+mgh$ $$E_B=mgH-mgh+mgh=mgH$$

3. At Position C (Ground, Height $0$):

• The body has fallen the full distance $H$.
• Let the velocity just before hitting the ground be $v_C$.
• Using $v^2=u^2+2as$: $$v_C^2=0^2+2gH=2gH$$
• $K{E}_C=\frac{1}{2}m{v}_C^2=\frac{1}{2}m(2gH)=mgH$
• $P{E}_C=mg(0)=0$
• Total Energy $E_C=K{E}_C+P{E}_C=mgH+0=mgH$

Conclusion: Comparing the total mechanical energy at all three positions: $$E_A=E_B=E_C=mgH$$

Since the sum of kinetic and potential energies remains constant ($mgH$) throughout the motion, the law of conservation of mechanical energy is proved.