A shell is fired at a speed of 200 m/s at an | vedclass

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(A) Let the tower be at the origin $(0, 80)$. The initial velocity of shell $1$ is $\vec{v}_1 = 200(\cos 37^o \hat{i} + \sin 37^o \hat{j}) = 200(0.8 \

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(A) Let the tower be at the origin $(0, 80)$. The initial velocity of shell $1$ is $\vec{v}_1 = 200(\cos 37^o \hat{i} + \sin 37^o \hat{j}) = 200(0.8 \hat{i} + 0.6 \hat{j}) = 160 \hat{i} + 120 \hat{j}\ m/s$.The jeep moves with $\vec{v}_J = 10 \hat{i}\ m/s$. The velocity of shell $2$ relative to the jeep is $\vec{v}_{2J} = 250(\cos 53^o \hat{i} + \sin 53^o \hat{j}) = 250(0.6 \hat{i} + 0.8 \hat{j}) = 150 \hat{i} + 200 \hat{j}\ m/s$.The absolute velocity of shell $2$ is $\vec{v}_2 = \vec{v}_{2J} + \vec{v}_J = (150 \hat{i} + 200 \hat{j}) + 10 \hat{i} = 160 \hat{i} + 200 \hat{j}\ m/s$.Since both shells are under gravity,their relative acceleration is $\vec{a}_{21} = \vec{a}_2 - \vec{a}_1 = (-g \hat{j}) - (-g \hat{j}) = 0$. Thus,the relative motion is uniform.The relative velocity is $\vec{v}_{21} = \vec{v}_2 - \vec{v}_1 = (160 \hat{i} + 200 \hat{j}) - (160 \hat{i} + 120 \hat{j}) = 80 \hat{j}\ m/s$.The initial relative position is $\vec{r}_{21} = \vec{r}_2 - \vec{r}_1 = (0, 0) - (0, 80) = -80 \hat{j}\ m$.The relative position at time $t$ is $\vec{r}(t) = \vec{r}_{21} + \vec{v}_{21} t = -80 \hat{j} + 80t \hat{j} = 80(t-1) \hat{j}$.The shells are closest when the magnitude of the relative position vector is minimum,which occurs at $t = 1\ s$,where the relative distance is $0$.

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