A ball is projected from ground with a speed of $35{\mathrm{ms}}^{-1}$ at an angle of $37°$ with the horizontal as shown in the figure.

A wall of height $10\mathrm{m}$ is there at a distance of $70\mathrm{m}$ from the point of projection as shown. Determine whether the ball will hit the wall or not. If the ball hits the wall, then determine where it will hit and if not then by how much it will fail to do so ? [Take $g=9.8{\mathrm{ms}}^{-2}$ ]

Step 1: Given data and approach

The height of the wall is $10\text{m}$.

The initial speed of the ball, $u=35{\mathrm{ms}}^{-1}$.

The angle of projectile, $\theta =37°$.

The distance of the wall from the point of projection, $x=70\mathrm{m}$.

Step 2: Find the vertical displacement of the ball

The equation of trajectory is $\mathit{y}\mathbf{=}\mathit{x}\mathbf{tan}\mathit{\theta }\mathbf{-}\frac{\mathbf{g}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{u}}^{\mathbf{2}}{\mathbf{cos}}^{\mathbf{2}}\mathbf{\theta }}$.

On putting the given values in the above equation, we get:

$$\begin{gathered} \Rightarrow y=70\tan 37°-\frac{9.8\times {70}^2}{2\times {35}^2{\cos }^{2}37°} \\ \Rightarrow y=22.019\text{m} \end{gathered}$$

The height of the wall is less than the vertical displacement of the ball. Therefore, the ball cannot hit the wall.

The ball will miss the wall by $\left(22.019-10\right)=12.019\text{m}$

Hence, the ball cannot hit the wall.