Question

4. A projectile is launched from a latitude 45 °N and travels 20 km to the north at 1000 m/s. a. What is the direction and magnitude of the Coriolis force and Coriolis acceleration? b. Calculate the lateral displacement due to the Coriolis force when the projectile reaches its target. Explain your reasoning. Recall that the displacement resulting from an acceleration a over time t is d = at/2.

Answer 1

a) here we have a projectile shot from 45° north latitude$(\phi)$,with a velocity V= 1000m/s

So the Coriolis force F_{​​​​​​c} is given as

Fe=22mV sin()

Here $\Omega$ is the rate of rotation of earth which is 7.26×10^-5rad/s

.:. F. = 2 * 7.26 x 10-5 x 1000 x sin(45) = 0.102N

The Coriolis acceleration a​​​_{​​​​​​c} =F_c/m is given as

ac = 2VI2sin(o)

ac = 2 x 1000 x 0.00014 x 0.70 = 0.20m/s2

Since the projectile is launched in northern hemisphere, it will deflect rightward i.e. towards eastward. So the direction of Coriolis force and Coriolis acceleration is eastward.

b) in order to determine the lateral displacement,we first need to calculate the time of flight of the projectile which is given as

202×1000 -=204.08sec g 9.8

We know that

a do = 2VI2sin(0)

Integrating above equation twice we get

X = V sin(o)t

Which we can clearly see,is same as d=at^{​​​​​2}​/2 because

a = 2VI2sin()

Therefore lateral displacement d is given as

d=VI2Sin(0)

..d = 1000 x 7.26 x 10-5 X sin(45) x (204.084) = 2140.58m

Therefore lateral displacement=2140..48 m