Kinematics question .

a pilot is flying from montreal, a displacement of 571 km [e 13.0(degrees) n]. the airplane is capable of flying at 301 km/h in still air. during the flight there is a constant wind of 95 km/h blowing form the direction of [e 30.0(degrees) n]. assuming she flies directly to montreal, (along the displacement vector) calculate the plane's speed relative to the ground and the direction the pilot should point the airplane to maintain his direct route to montreal.

The velocity of the air relative to the ground is

The velocity of the plane relative to the air is

where is the direction the plane needs to point so that the resultant vector - the velocity of the plane relative to the ground - is

Here is the speed of the plane relative to the ground.

So we have

Use a calculator to solve for and ; you should find

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