Two objects are located at a height of 10 m | vedclass

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(B) Given: Initial velocity $U_1 = U_2 = 2 \sqrt{2} \ m \ s^{-1}$, acceleration due to gravity $g = 10 \ m \ s^{-2}$.At time $t$, the velocity vectors

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$0.4$

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(B) Given: Initial velocity $U_1 = U_2 = 2 \sqrt{2} \ m \ s^{-1}$, acceleration due to gravity $g = 10 \ m \ s^{-2}$.At time $t$, the velocity vectors are given by:$\vec{v}_1 = (U_1 \cos 45^{\circ}) \hat{i} + (U_1 \sin 45^{\circ} - gt) \hat{j} = (2 \sqrt{2} \cdot \frac{1}{\sqrt{2}}) \hat{i} + (2 \sqrt{2} \cdot \frac{1}{\sqrt{2}} - 10t) \hat{j} = 2 \hat{i} + (2 - 10t) \hat{j} \dots(1)$$\vec{v}_2 = (U_2 \cos 135^{\circ}) \hat{i} + (U_2 \sin 135^{\circ} - gt) \hat{j} = (2 \sqrt{2} \cdot -\frac{1}{\sqrt{2}}) \hat{i} + (2 \sqrt{2} \cdot \frac{1}{\sqrt{2}} - 10t) \hat{j} = -2 \hat{i} + (2 - 10t) \hat{j} \dots(2)$Since the velocity vectors are perpendicular, their dot product must be zero:$\vec{v}_1 \cdot \vec{v}_2 = 0$$(2 \hat{i} + (2 - 10t) \hat{j}) \cdot (-2 \hat{i} + (2 - 10t) \hat{j}) = 0$$-4 + (2 - 10t)^2 = 0$$(2 - 10t)^2 = 4$Taking the square root on both sides:$2 - 10t = \pm 2$Case $1$: $2 - 10t = 2 \Rightarrow 10t = 0 \Rightarrow t = 0 \ s$Case $2$: $2 - 10t = -2 \Rightarrow 10t = 4 \Rightarrow t = 0.4 \ s$Thus, the velocity vectors are perpendicular at $t = 0.4 \ s$.

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