Two airplanes leave an airport at the same time and travel in opposite directions. one plane travels 87 km/h faster than the other. if the two planes are 11,865 km apart after 7 hours, what is the rate of each plane?

Speed of A = 891 km/h

Speed of B = 804 km/h

Explanation:

Let the speed of aeroplane B is v, the speed of aeroplane is A is 87 km/h faster than B.

So, the speed of aeroplane A is 87 + v.

Distance traveled by A after 7 hours, d1 = (87 + v) x 7

Distance traveled by B after 7 hours, d2 = v x 7

Total distance traveled = 11865 km

So, d1 + d2 = 11865

(87 + v) x 7 + v x 7 = 11865

609 + 14 v = 11865

14 v = 11256

v = 804 km/h

So, the speed of A = 87 + 804 = 891 km/h

Speed of B = 804 km/h

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initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft