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lashaaungas
06.11.2020 •
Physics
A B-52 bomber jet flies at a horizontal velocity of 286.2 m/s and at an altitude of 7500 m above the ground. How far away horizontally should a payload be dropped to land on a target?
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Ответ:
11196.14 m
Explanation:
From the question given above, the following data were obtained:
Horizontal velocity (u) = 286.2 m/s
Height (h) = 7500 m
Horizontal distance (s) =?
Next, we shall determine the time taken for the payload to get to the target. This can be obtained as follow:
Height (h) = 7500 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
7500 = ½ × 9.8 × t²
7500 = 4.9 × t²
Divide both side by 4.9
t² = 7500 / 4.9
Take the square root of both side
t = √(7500 / 4.9)
t = 39.12 s
Finally, we shall determine the horizontal distance as follow:
Horizontal velocity (u) = 286.2 m/s
Time (t) = 39.12 s
Horizontal distance (s) =.?
s = ut
s = 286.2 × 39.12
s = 11196.14 m
Thus the payload will travel 11196.14 m horizontally in order to hit the target
Ответ:
v = u - gt
v = 0 - 9.8(1.75)
v = - 17.15 m/s