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## Q. 6.9

Percy Panther is snoozing in an open-top artillery truck when he senses the presence of the mischievous Arnold Aardvark lurking beneath. He quietly releases the handbrake to escape down the hill inclined at an angle α.

Arnold Aardvark having quietly rigged a remote trigger, immediately fires the gun, launching a shell of mass m straight up from the truck, as shown in Fig. 6.4. The gun has a muzzle velocity  $\mathbf{v}_0$,  and the total mass of the truck and its strange driver is M. Determine the time and the location at which the shell impacts the ground, and find the location of Percy Panther at that time. ## Verified Solution

First, we determine the motion of the shell S, whose free body diagram is shown in Fig. 6.4. The total force acting on S is its weight  $\mathbf{W}_S=m \mathrm{~g}$.  Thus, in the inertial frame  $\Phi=\left\{F ; \mathbf{i}_k\right\}$  fixed in the ground , the constant force in (6.22) and (6.23) is  $\mathbf{F}_0=\mathbf{W}_s=m g(\sin \alpha \mathbf{i}-\cos \alpha \mathbf{j})$;  and with  $\mathbf{v}_0=v_0 \mathbf{j}$  and  $\mathbf{x}_0=\mathbf{0}$  initially, we obtain, in evident notation ,

$\mathbf{v}(P, t)=\frac{\mathbf{F}_0}{m} t + \mathbf{v}_0$               (6.22)

$\mathbf{x}(P, t)=\frac{\mathbf{F}_0}{2 m} t^2 + \mathbf{v}_0 t + \mathbf{x}_0$                 (6.23)

$\mathbf{v}_S(t)=v_0 \mathbf{j} + g t(\sin \alpha \mathbf{i} – \cos \alpha \mathbf{j})$,                 (6.26a)

$\mathbf{x}_S(t)=\frac{1}{2} g t^2 \sin \alpha \mathbf{i} + \left(v_0 t – \frac{1}{2} g t^2 \cos \alpha\right) \mathbf{j}$.                   (6.26b)

Let the reader derive these results starting from (6.1), determine the maximum height reached by S, and show that its trajectory is a parabola.

$\mathbf{F}=\dot{\mathbf{p}}=m \mathbf{a}=m \dot{\mathbf{v}}=m \ddot{\mathbf{x}}$      (6.1)

The shell returns to the ground after a time t* when $\mathbf{x}_S\left(t^*\right)=r \mathbf{i}$   in Fig. 6.4, and hence by (6.26b) ,

$r=\frac{1}{2} g t^{* 2} \sin \alpha, \quad t^*=\frac{2 v_0}{g \cos \alpha} .$             (6.26c)

The results are independent of the mass or any other property of the shell. Elimination of t* from the first of (6.26c) yields the impact range r in terms of the muzzle speed $v_0$  and the angle α that the gun makes with the vertical axis of g:

$r=\frac{2 v_0^2 \tan \alpha}{g \cos \alpha}$.                  (6.26d)

Now consider the free body diagram of the truck in Fig. 6.4. The total force  $\mathbf{F}_T$   acting on the truck is its total weight  $\boldsymbol{W}_T$  and the normal surface reaction force N. Without frictional effects, (6.1) becomes

$\mathbf{F}_T=\mathbf{N} + \mathbf{W}_T=N \mathbf{j} + M g(\sin \alpha \mathbf{i} – \cos \alpha \mathbf{j})=M \mathbf{a}_T$.            (6.26e)

Since the truck accelerates along the i direction,  $N=M g \cos \alpha \text { and } \mathbf{a}_T=g \sin \alpha \mathbf{i} .$  Hence, two easy  integrations with  $\mathbf{v}_0=\mathbf{0} \text { and } \mathbf{x}_0=\mathbf{0}$  yield

$\mathbf{v}_T(t)=g t \sin \alpha \mathbf{i}$              (6.26f)

$\mathbf{x}_T(t)=\frac{1}{2} g t^2 \sin \alpha \mathbf{i}$                   (6.26g)

Comparison of the i components in (6.26b) and (6.26g) parallel to the truck’s motion reveals that the shell at each instant is directly above the truck, now  coasting toward the ultimate surprise! But a few tenths of a second before the impending catastrophe, Percy Panther spots the converging shell and slams on the brakes. The shell explodes violently in front of the truck, destroying it. Through the smoky haze, Arnold Aardvark spies the black, whisker-singed and disheveled driver crawling safely away to seek revenge another day.