91Ó°ÊÓ

It is given that,

The initial velocity ofhuman cannonball$v_0=26.5\mathrm{m}/\mathrm{s}$

The angle with which the human cannonball is launched$\mathrm{θ}=53°$

The problem deals with the projectile path of an object. Any object thrown into the sky will end up falling because of the action of gravity. The direction followed by it from the beginning of the motion until it falls back is its path which is in the shape of a parabola. In short the path of the trajectory is parabola. Here, using the kinematic equations and equation for the projectile path,the clearance of the ball over the wheels and the position of the center of the net from the canon can be found.

Formulae:

The equation for projectile path is given by,

$∆y=x\tan {\mathrm{θ}}_0-\frac{g{x}^2}{2{(v_0\cos \mathrm{θ})}^2}$ (i)

Where

$∆\mathrm{y}=y-y_0$

The Newton’s kinematic equation is given by,

$v_f^2=v_0^2+2a\left(∆y\right)$ (ii)

The equation for projectile path is,

$$\begin{gathered} y=y_0+x\tan {\mathrm{θ}}_0-\frac{g{x}^2}{2{(v_0\cos {\mathrm{θ}}_0)}^2} \\ y=3+23\tan \left(53\right)-\frac{9.8×{23}^2}{2{\left(26.5\cos \left(53\right)\right)}^2} \\ 3+30.52-10.19 \\ y=23.328 \\ ~23.3m \end{gathered}$$

As the height of the wheel is 18 m, the clearance of the human cannonball over the first wheel is,

$$\begin{gathered} \mathrm{h}=23.3-18 \\ =5.3\mathrm{m} \end{gathered}$$

Over the middle wheel, the human cannonball reaches the maximum height. So, at that point, it has zero velocity.

Using equation (ii),

$$\begin{gathered} 0={\left(26.5×\sin \left(53\right)\right)}^2-2×9.8×\left(y-y_0\right) \\ 0={\left(26.5×\sin \left(53\right)\right)}^2-2×9.8×\left(y-3\right) \\ y=25.85m \end{gathered}$$

As the height of the wheel is 18 m, the clearance of the human cannonball over the middle wheel is,

$\begin{array}{l}h=25.85-18\\ =7.85\\ ≈7.9m\end{array}$

The total time of flight for the motion of the human cannonball is,

$2x2.15=4.3s$

The horizontal displacement of thehuman cannonball is,

$$\begin{gathered} x=v_0\cos \mathrm{θ}×t \\ x=26.5\cos \left(53\right)(4.3) \\ x=68.57 \\ ≈69m \end{gathered}$$

Therefore, the position of the center of the net from the canon is 69 m