A projectile is thrown from a point O on the | vedclass

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Question

(A) Initial velocity $u = 5\sqrt{2} 	ext{ m/s}$ at $45^{\circ}$ from the vertical means it is also at $45^{\circ}$ from the horizontal. Thus,$u_x = u

Vedclass · Accepted Answer

$0.5, 7.5$

Vedclass · Answer

(A) Initial velocity $u = 5\sqrt{2} 	ext{ m/s}$ at $45^{\circ}$ from the vertical means it is also at $45^{\circ}$ from the horizontal. Thus,$u_x = u \cos 45^{\circ} = 5\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 5 	ext{ m/s}$ and $u_y = u \sin 45^{\circ} = 5\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 5 	ext{ m/s}$.Range $R = \frac{2 u_x u_y}{g} = \frac{2 	imes 5 	imes 5}{10} = 5 	ext{ m}$.Time of flight $T = \frac{2 u_y}{g} = \frac{2 	imes 5}{10} = 1 	ext{ s}$.The projectile splits at the highest point,which occurs at time $T/2 = 0.5 	ext{ s}$ and horizontal distance $R/2 = 2.5 	ext{ m}$.One part falls vertically,meaning its horizontal velocity becomes $0$. It takes $0.5 	ext{ s}$ to reach the ground,so $t = 0.5 	ext{ s}$.By conservation of linear momentum in the horizontal direction: $M u_x = (M/2) v_1 + (M/2) v_2$. Since the first part falls vertically,$v_1 = 0$. Thus,$M(5) = (M/2) v_2 \Rightarrow v_2 = 10 	ext{ m/s}$.The second part travels horizontally with velocity $10 	ext{ m/s}$ for $t = 0.5 	ext{ s}$ from the position $R/2 = 2.5 	ext{ m}$.Horizontal distance covered by the second part after splitting $= v_2 	imes t = 10 	imes 0.5 = 5 	ext{ m}$.Total distance $x$ from $O = (R/2) + 5 = 2.5 + 5 = 7.5 	ext{ m}$.Therefore,$t = 0.5 	ext{ s}$ and $x = 7.5 	ext{ m}$.

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