Achild walks 5.0 meters north, then 4.0 meters east, and finally 2.0 meters south. what is the magnitude of the resultant displacement of the child after the entire walk

Resultant displacement of the child is 5 m.

Explanation:

It is given that, a child walks 5.0 meters north, then 4.0 meters east, and finally 2.0 meters south.

We have to find the magnitude of displacement of the child after the entire walk. The attached figure shows entire journey of the child. Displacement of an object is equal to the shortest path covered by it. Here, AM shows the net displacement. It can be calculated using Pythagoras theorem as :

Since, AB = 5 m So, OA = 3 m and BC = OM = 4 m

⇒ AM = 5 m

Hence, the resultant displacement of the child after the entire walk is 5 m.

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