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(A) The horizontal component of velocity remains constant throughout the flight because there is no horizontal force (ignoring air resistance).Let $V_

Vedclass · Accepted Answer

$\frac{1}{2}(\sqrt{3} + 1) mV$

Vedclass · Answer

(A) The horizontal component of velocity remains constant throughout the flight because there is no horizontal force (ignoring air resistance).Let $V_1$ be the final speed at the ground.$V \cos 30^{\circ} = V_1 \cos 45^{\circ}$$V \frac{\sqrt{3}}{2} = V_1 \frac{1}{\sqrt{2}}$$V_1 = V \sqrt{\frac{3}{2}}$Impulse $J$ is equal to the change in momentum: $\vec{J} = \vec{p}_f - \vec{p}_i$.Since horizontal momentum is constant,the impulse acts only in the vertical direction.$J = |p_{fy} - p_{iy}| = |m(-V_1 \sin 45^{\circ}) - m(V \sin 30^{\circ})|$$J = m |V \sqrt{\frac{3}{2}} \cdot \frac{1}{\sqrt{2}} + V \cdot \frac{1}{2}|$$J = m |V \frac{\sqrt{3}}{2} + \frac{V}{2}| = \frac{1}{2}(\sqrt{3} + 1) mV$.

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