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?

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

this is poiselle's equation from fluid mechanic in physics. mech eng

c. magnitude and direction