A ball is thrown from ground level so as to just clear a wall 4 metres high at a distance of 4 metres and falls at a distance of 14 metres behind the wall. The magnitude velocity of projection of the ball will be:

Referring to (fig.) let P be a point on the trajectory whose co-ordinates are (4, 4). As the ball strikes the ground at a distance 14 metre from the wall, the range is 4 + 14 = 18 metre. The equation of trajectory is

$\text{y}=\text{x tan}\theta -\left(\frac{g}{2}\right) \frac{x^2}{u^2{\cos }^2\theta }$

$\begin{array}{l}\text{ or }\mathrm{y}=\mathrm{xtan}\theta \left[1-\frac{gx}{2u^2{\cos }^2\theta \cdot \tan \theta }\right]\\ \text{ or }\mathrm{y}=\mathrm{xtan}\theta \left[1-\frac{g}{2u^2}\frac{x}{\sin \theta \cos \theta }\right]\\ y=\mathrm{xtan}\theta \left[1-\frac{x}{R}\right]; where R is range\end{array}$

$\begin{array}{l}\text{ here }x=4,y=4\text{ and }R=18\\ \therefore 4=4\tan \theta \left[1-\frac{4}{18}\right]=4\tan \theta \left(\frac{7}{9}\right)\\ \text{ or }\tan \theta =\frac{9}{7};\sin \theta =\frac{9}{\sqrt{130}}\text{ and }\cos \theta =\frac{7}{\sqrt{130}}\\ \text{ Again }R=\frac{2}{g}{u}^2\sin \theta \cos \theta ; in this equation substitute above obtained values\end{array}$

$$\begin{gathered} range R=\left(\frac{2}{9.8}\right)\times {\mathrm{u}}^2\times \left(\frac{9}{\sqrt{130}}\right)\times \left(\frac{7}{\sqrt{130}}\right) but R=18 \\ \Rightarrow {\text{u}}^{\text{2}} =\frac{18\times 9.8\times \sqrt{130}\times \sqrt{130}}{2\times 9\times 7} =\frac{98\times 13}{7}=\text{ 182}, \\ \Rightarrow \text{u }=\sqrt{182}Metre per second. \end{gathered}$$