A ball is projected at 60^{circ} from the hor | vedclass

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(B) The initial velocity of the ball is $u = 200 \, m/s$ at an angle $	heta = 60^{\circ}$.At the maximum height,the vertical component of velocity is

Vedclass · Accepted Answer

$300 \, m/s$ horizontal

Vedclass · Answer

(B) The initial velocity of the ball is $u = 200 \, m/s$ at an angle $	heta = 60^{\circ}$.At the maximum height,the vertical component of velocity is $0$,and the horizontal component is $v_x = u \cos 60^{\circ} = 200 	imes 0.5 = 100 \, m/s$.The total mass of the ball is $m$. At maximum height,the momentum is $\vec{P}_{initial} = m 	imes 100 \hat{i}$.The ball explodes into $3$ equal fragments of mass $m/3$ each.Let the velocities of the fragments be $\vec{v}_1 = 100 \hat{j}$,$\vec{v}_2 = -100 \hat{j}$,and $\vec{v}_3 = \vec{V}$.By the law of conservation of linear momentum:$\vec{P}_{initial} = \vec{P}_{final}$$m(100 \hat{i}) = \frac{m}{3}(100 \hat{j}) + \frac{m}{3}(-100 \hat{j}) + \frac{m}{3}(\vec{V})$$100 \hat{i} = \frac{1}{3}(100 \hat{j} - 100 \hat{j} + \vec{V})$$100 \hat{i} = \frac{1}{3} \vec{V}$$\vec{V} = 300 \hat{i} \, m/s$.Thus,the third part moves at $300 \, m/s$ in the horizontal direction.

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