Question

In 1939 or 1940, Emanuel Zacchini took his human cannonball act to an extreme: After being shot from a cannon, he soared over three Ferris wheels and into a net (Fig.4-39). Assume that he is launched with a speed of 26.5 m/sand at an angle of $\mathbf{53}\mathbf{.}\mathbf{0}\mathbf{°}$. (a)Treating him as a particle, calculate his clearance over the first wheel. (b) If he reached maximum height over the middle wheel, by how much did he clear it? (c) How far from the cannon should the net’s center have been positioned (neglect air drag)?

Short answer

1. The clearance of the human cannonball over the first wheel is 5.3 m.
2. The clearance of the human cannonball over the middle wheel is 7.9 m.
3. The position of the center of the net from the canon is 69 m.

Step-by-step solution

Given information

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$\theta =53°$

Determining the concept

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,

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

(a) Determining the clearance of the human cannonball over the first wheel

The equation for projectile path is,

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

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

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

(b) Determining the clearance of the human cannonball over the middle wheel

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

Using equation (ii),

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

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

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

(c) Determining the position of the center of the net from the canon

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

$2x2.15=4.3s$

The horizontal displacement of thehuman cannonball is,

$$\begin{gathered} x=v_0\cos \theta \times t \\ x=26.5\cos \left(53\right)(4.3) \\ x=68.57 \\ \approx 69m \end{gathered}$$

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