A bomb is dropped by an airplane flying horiz | vedclass

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(C) The airplane is flying horizontally,so the initial vertical component of the bomb's velocity is $u_y = 0 	ext{ m/s}$.Using the equation of motion

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$\frac{10^4}{9 \sqrt{2}} 	ext{ m}$

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(C) The airplane is flying horizontally,so the initial vertical component of the bomb's velocity is $u_y = 0 	ext{ m/s}$.Using the equation of motion for the vertical direction: $h = \frac{1}{2} gt^2$.Substituting the given values: $980 = \frac{1}{2} 	imes 9.8 	imes t^2$.$t^2 = \frac{980 	imes 2}{9.8} = 100 	imes 2 = 200$.$t = \sqrt{200} = 10\sqrt{2} 	ext{ s}$.The horizontal velocity of the bomb is $v_x = 200 	ext{ km/hr} = 200 	imes \frac{5}{18} = \frac{1000}{18} 	ext{ m/s}$.The horizontal distance $d$ covered by the bomb during its fall is $d = v_x 	imes t$.$d = \frac{1000}{18} 	imes 10\sqrt{2} = \frac{10000\sqrt{2}}{18} = \frac{5000\sqrt{2}}{9} 	ext{ m}$.Wait,simplifying the expression: $d = \frac{1000}{18} 	imes 10\sqrt{2} = \frac{10000}{9\sqrt{2}} 	ext{ m}$.

Column-$I$	Column-$II$
$(1)$ Angle of projection for a projectile launched horizontally with constant speed	$(a)$ $0$
$(2)$ Horizontal component of acceleration for a projectile launched horizontally with constant speed	$(b)$ $0^o$

$A$ bomb is dropped by an airplane flying horizontally with a velocity of $200 \text{ km/hr}$ at a height of $980 \text{ m}$. At the time of dropping the bomb,the horizontal distance of the airplane from the target on the ground to hit it directly is (given $g = 9.8 \text{ m/s}^2$):

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Match Column-$I$ with Column-$II$.
Column-$I$ Column-$II$
$(1)$ Angle of projection for a projectile launched horizontally with constant speed $(a)$ $0$
$(2)$ Horizontal component of acceleration for a projectile launched horizontally with constant speed $(b)$ $0^o$

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