The horizontal distance XQ traveled by the bomb is 250 m.
Distance X Q
Let the XQ be the horizontal distance traveled by the bomb.
Time for the bomb to drop from 100 m
h = vt + ¹/₂gt
Let the vertical velocity = 0
h = 0 + ¹/₂gt²
h = ¹/₂gt²
2h = gt²
t² = 2h/g
t = √(2h/g)
t = √(2 x 100 / 9.8)
t = 4.5 s
Horizontal distance traveled by the bomb
XQ = vx(t)
where;
vx is horizontal speed, = 200 km/hr = 55.56 m/s
XQ = 55.56 x 4.5
XQ = 250 m
Thus, the horizontal distance XQ traveled by the bomb is 250 m.
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