Question

A boy kicks a football horizontally from the roof of a building of a height of $10m$. If the line joins the initial position of the football and the point where it hits the ground make an angle of $45°$ with the ground, then find the displacement of the football

Answer 1

Step 1: Given data

The diagrammatic representation of the given data is given below:

Height of the building, $OA=10m$.

Angle of the football with the ground on reaching the ground, $\angle ABO=45°$

Step 2: Find the distance $\mathit{O}\mathit{B}$ from the building at which the football touches the ground

As we know, $\tan \theta =\frac{Perpendicular}{Base}$

In the right-angled triangle $AOB$.

$\tan 45°=\frac{OA}{OB}$

$$\begin{gathered} 1=\frac{10}{OB} \\ OB=10m \end{gathered}$$

Step 3: Find the net displacement $\mathit{A}\mathit{B}$ of the football.

Use the Pythagoras theorem in the right-angled triangle $AOB$.

$\mathit{H}\mathit{y}\mathit{p}\mathit{o}\mathit{t}\mathit{e}\mathit{n}\mathit{o}\mathit{u}\mathit{s}\mathbf{=}\sqrt{{\mathbf{P}\mathbf{e}\mathbf{r}\mathbf{p}\mathbf{e}\mathbf{n}\mathbf{d}\mathbf{i}\mathbf{c}\mathbf{u}\mathbf{l}\mathbf{a}\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{B}\mathbf{a}\mathbf{s}\mathbf{e}}^{\mathbf{2}}}$

$$\begin{gathered} AB=\sqrt{O{A}^2+O{B}^2} \\ AB=\sqrt{{10}^2+{10}^2} \\ AB=\sqrt{100+100} \\ AB=\sqrt{200} \\ AB=10\sqrt{2}m \end{gathered}$$

Hence, the net displacement of the football is $\mathbf{10}\sqrt{\mathbf{2}}\mathbf{}\mathit{m}$.