A projectile is projected at a speed of $75{\text{ms}}^{-1}$ from ground level, at an angle of $60°$ above the horizontal. A wall of height $11\mathrm{m}$ is located at a distance of $27\mathrm{m}$ from the point of projection. Determine by how much amount the projectile will clear the top of the wall. [Take $g=10{\mathrm{ms}}^{-2}$ ]

Step 1: Given data

The speed of the projectile is $u=75{\text{ms}}^{-1}$.

The angle of projection is $\theta =60°$.

The wall is at a distance of $27\mathrm{m}$ from the point of projection.

The acceleration of gravity is $g=10{\mathrm{ms}}^{-2}$.

Therefore, assume the horizontal displacement of the projectile as $x=27\mathrm{m}$.

Step 2: Calculate the time taken from the equation of the horizontal displacement

Consider the time taken as $t$ seconds and find it from the equation of the horizontal displacement.

The equation of the horizontal displacement is given as $\mathit{x}\mathbf{=}\mathit{u}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{\times }\mathit{t}$

$$\begin{gathered} \Rightarrow 27=75cos60°\times t \\ \Rightarrow t=\frac{27\times 2}{75} \\ \Rightarrow t=\frac{18}{25}s \end{gathered}$$

Step 3: Calculate the vertical displacement and the amount by which the projectile clears the top of the wall

The equation of the vertical displacement is given as $\mathit{y}\mathbf{=}\mathit{u}\mathbf{sin}\mathit{\theta }\mathbf{\times }\mathit{t}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{g}{\mathit{t}}^{\mathbf{2}}$

$$\begin{gathered} \Rightarrow y=(75\sin 60°\times \frac{18}{25})-(\frac{1}{2}\times 10\times \frac{{18}^2}{{25}^2}) \\ \Rightarrow y=(3\times \frac{\sqrt{3}}{2}\times 18)-\left(\frac{1620}{{25}^2}\right) \\ \Rightarrow y=44.172m \end{gathered}$$

Step 4: Calculate the vertical displacement and the amount by which the projectile clears the top of the wall

Subtract the height of the wall $11\mathrm{m}$ from the vertical displacement $y$ of the projectile to get the required answer.

Therefore, the amount by which the projectile clears the top of the wall is:

$$\begin{gathered} \Rightarrow y-11 \\ \Rightarrow 44.173-11 \\ =33.173\text{m} \end{gathered}$$

Hence, the amount by which the projectile clears the top of the wall is $\mathbf{33}\mathbf{.}\mathbf{173}\mathbf{}\mathbf{m}$.