Question

The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon, using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon, and at a range of

69 m

. He was propelled inside the barrel for

5.2 m

and launched at an angle of

53 °

. If his mass was

85 kg

and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at

53 °

. Neglect the air drag.)

Accepted Answer

The shot by Emanuel Zucchini over Ferris wheels to land is like a projectile motion. The range of the projectile is,

$R = \frac{u^{2} \sin 2 θ}{g}$ ...(1)

The range is,

$R = 69 m$

The angle of projection is,

$θ = 53 °$

From (1),

$u = (\sqrt{\frac{R g}{\sin 2 θ}})$

$u = (\sqrt{\frac{(69 m) (9.8 m s^{- 2})}{\sin 2 (53 °)}}) = 26.52 m s^{- 1}$

The projectile motion is a two-dimensional motion. The vertical and horizontal components of velocity are,

$u_{x} = u \cos θ = 26.52 m s^{- 1} (\cos (53 °)) = 15.96 m s^{- 1} u_{y} = u \sin θ = 26.52 m s^{- 1} (\sin (53 °)) = 22.18 m s^{- 1}$

The person moves with a constant acceleration in the horizontal direction. The acceleration is,

$a_{x} = \frac{{u_{x}}^{2}}{2 x}$

$a_{x} = \frac{{(15.96 m s^{- 1})}^{2}}{2 (5.2 m) \cos (53 °)} = 40.7 m s^{- 2}$

The component of the force along the horizontal direction, where mass $m = 85 kg$ ,

$F_{x} = m a_{x} = (85 kg) (40.7 m s^{- 2}) = 3460 N$

The component of the acceleration in the vertical direction,

$a_{y} = \frac{{u_{y}}^{2}}{2 y} = \frac{(21.18 m s^{- 2})}{2 (5.2) \sin (53 °)} = 54 m s^{- 2}$

The component of the force along the vertical direction,

$F_{y} = m a_{y} + m g = (85 kg) (54 m s^{- 2} + 9.8 m s^{- 2}) = 5424 N$

The magnitude of the force,

$F = \sqrt{{F_{x}}^{2} + {F_{y}}^{2}} = \sqrt{{(3600)}^{2} + {(5424)}^{2}} = 6.4 \times 10^{3} N$

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