The initial speed of a bullet fired from a ri | vedclass

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(C) Let $t$ be the time taken by the bullet to hit the target.Horizontal distance $d = 700 \; m$ and horizontal velocity $v_x = 630 \; m/s$.Since ther

Vedclass · Accepted Answer

$6.17$

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(C) Let $t$ be the time taken by the bullet to hit the target.Horizontal distance $d = 700 \; m$ and horizontal velocity $v_x = 630 \; m/s$.Since there is no horizontal acceleration,$t = \frac{d}{v_x} = \frac{700}{630} = \frac{10}{9} \; s$.For vertical motion,the bullet is subject to gravity. The vertical displacement $h$ (drop) is given by $h = u_y t + \frac{1}{2} g t^2$.Since the initial vertical velocity $u_y = 0$,we have $h = \frac{1}{2} g t^2$.Substituting the values: $h = \frac{1}{2} 	imes 10 	imes \left( \frac{10}{9} ight)^2$.$h = 5 	imes \frac{100}{81} = \frac{500}{81} \approx 6.17 \; m$.Therefore,the rifle must be aimed $6.17 \; m$ above the center of the target to compensate for the gravitational drop.

The initial speed of a bullet fired from a rifle is $630 \; m/s$. The rifle is fired at the center of a target $700 \; m$ away at the same level as the target. How far above the center of the target (in $m$) must the rifle be aimed in order to hit the target? (Take $g = 10 \; m/s^2$)

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