If the jet is moving at a speed of 1250 km/h at the lowest point of the loop, determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.5 g's.

The minimum radius of the circle is 1892.75 meters

Explanation:

Speed of the jet, v = 1250 km/h = 347.23 m/s

We need to find the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.5 g.

The centripetal acceleration is given by :

So, the minimum radius of the circle is 1892.75 meters. Hence, this is the required solution.                                                          

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

explanation:

to push or pull an object in a certain direction is known as force. a force in moving an object from one place to another.

force depends on the mass of an object. so, more is the mass of an object more will be the force required to move the object.

thus, we can conclude that the use of force to move an object is push or pull.