Givens:

The Distance from the origin to start point is: \(\displaystyle{y}_{{{0}}}={1.22}{m}\)

The launch angle is: \(\displaystyle\theta_{{{0}}}={45}^{{\circ}}\)

The Horizontal rangle is: \(\displaystyle{R}={107}{m}\)

Step 2

Part a

For 7.32 m high fence, the Horizontal distance from the origin is:

\(\displaystyle{x}-{x}_{{{0}}}={97.5}{m}\)

To determine the clearance over the high fence, we use

\(\displaystyle\Delta{y}={y}-{h}\)

So firstly, we need to find the maximum height:

\(\displaystyle{y}-{y}_{{{0}}}={\left({\tan{\theta}}_{{{0}}}\right)}{x}-{\frac{{{g}{x}^{{{2}}}}}{{{2}{\left({v}_{{{0}}}{\cos{\theta}}_{{{0}}}\right)}^{{{2}}}}}}\)

We will use the Horizontal rangle:

\(\displaystyle{R}={\frac{{{{v}_{{{0}}}^{{{2}}}}}}{{{g}}}}{\sin{{2}}}\theta_{{{0}}}\)

\(\displaystyle{107}{m}={\frac{{{{v}_{{{0}}}^{{{2}}}}}}{{{9.8}{\frac{{{m}}}{{{s}^{{{2}}}}}}}}}{\sin{{\left({2}\times{45}^{{\circ}}\right)}}}\)

\(\displaystyle{v}_{{{0}}}={32.4}{\frac{{{m}}}{{{s}}}}\)

Step 3

Then,

\(\displaystyle{y}={1.22}{m}+{\left({\tan{{45}}}^{{\circ}}\right)}{\left({97.5}\right)}-{\frac{{{\left({9.8}{\frac{{{m}}}{{{s}^{{{2}}}}}}\right)}{\left({97.5}{m}\right)}^{{{2}}}}}{{{2}{\left({32.4}{\frac{{{m}}}{{{s}}}}{\cos{{45}}}^{{\circ}}\right)}^{{{2}}}}}}\)

\(\displaystyle={10}{m}\)

The clearance of the ball can be obtained from:

\(\displaystyle\Delta{y}={y}-{h}\)

\(\displaystyle={10}{m}-{7.32}{m}\)

\(\displaystyle={2.7}{m}\)

Yes, the ball clear above the fence.

Step 4

Part b:

The center of the ball at: 10 m

The height of the fence is: 7.32 m

Then, the distance between the ball center and the fence top is: (clearance of the ball)

\(\displaystyle\Delta{y}={y}-{h}\)

\(\displaystyle={10}{m}-{7.32}{m}\)

\(\displaystyle\Delta{y}={2.7}{m}\)