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## Q. 6.10

Arnold Aardvark is sunbathing on a lookout platform at $\mathbf{x}_0=a \mathbf{i}+b \mathbf{j}$  in the frame $\Phi=\left\{O ; \mathbf{i}_k\right\}$  when he spots Percy Panther at O preparing to fire an artillery gun pointed directly toward the platform, as shown in Fig. 6.5. The gun has a muzzle velocity  $v_0$  and the tower is well within its range r . At the moment the gun is fired, Arnold Aardvark, sensing impending danger, grabs his umbrella, steps through a hole in the platform, and falls freely in pursuit of safety toward the ground. Determine the distance d that separates Arnold Aardvark and the shell at the instant t* when it crosses his line of fall. ## Verified Solution

The free body diagrams of the shell S and Arnold Aardvark B are shown in Fig . 6.5, in which $\mathbf{W}_S=m_S \mathbf{g} \text { and } \mathbf{W}_B=m_B \mathbf{g}$  denote their respective weights. Their free fall equations of motion, in evident notation, are

$\mathbf{F}_B=m_B \mathbf{g}=m_B \mathbf{a}_B, \quad \mathbf{F}_S=m_S \mathbf{g}=m_S \mathbf{a}_S$                      (6.27a)

Therefore, B and S have the same constant, free fall acceleration,  $\mathbf{a}_B=\mathbf{a}_S=\mathbf{g}$,  but their respective initial conditions differ. Integration of this equation, i.e.  $d \mathbf{v}_B=d \mathbf{v}_S$,  with  $\mathbf{v}_B(0)=\mathbf{0}$  and  $\mathbf{v}_S(0)=\mathbf{v}_0$,  the muzzle velocity of the gun, gives

$\mathbf{v}_B=\mathbf{v}_S-\mathbf{v}_0 \quad \text { with } \quad \mathbf{v}_0=v_0(\cos \beta \mathbf{i} + \sin \beta \mathbf{j})$                 (6.27b)

A second integration with $\mathbf{x}_B(0)=\mathbf{x}_0$  and $\mathbf{x}_S(0)=\mathbf{0}$  yields the relative position vector $\mathbf{D} \equiv \mathbf{x}_B-\mathbf{x}_S$  of B from S at any time t :

$\mathbf{D}=\mathbf{x}_0 – \mathbf{v}_0 t \quad \text { with } \quad \mathbf{x}_0=a \mathbf{i} + b \mathbf{j}$                  (6.27c)

At the instant t* when the shell crosses Arnold Aardvark’s line of escape  $x=a.  Thus, with $\mathbf{v}_0$  given in (6.27b) , (6.27c) yields  $d \mathbf{j}=\left(a – v_0 t^* \cos \beta\right) \mathbf{i} + \left(b – v_0 t^* \sin \beta\right) \mathbf{j}$.  The i component determines t*, and the j component yields

$d=b – a \tan \beta$               (6.27d)

for the distance separating Arnold Aardvark and the unyielding shell at t* , But Percy Panther had directed the gun on the line toward the platform with  $\tan \beta=b / a$;   so, Arnold Aardvark is headed straight toward an unpleasant surprise at the instant t*! But a few moments before disaster strikes, he spies the approaching shell and quickly fixes the crook-handled umbrella to a tower beam, instantly arresting his fall. The shell explodes violently beneath him, destroying the tower. Arnold Aardvark, his snout scorched and twisted , escapes the assault with renewed mischief in mind.

So long as the tower is within the gun’s range, the result is independent of the muzzle speed and of the masses of the objects involved; it depends only on the initial coordinates of B and the angle of elevation of the gun. Explain why  Arnold Aardvark, living in a world where this solution is meaningful, was wise not to have used the umbrella as a parachute.