Question

An object is dropped out of an airplane that is moving horizontally at 350 m/s and is 22,000 m above the ground. Ignoring friction, what will its approximate VF X be on impact?

Answer 1

Vx = 350m/s

Vy = 656.99 m/s

Step-by-step explanation:

at the time of drop, the horizontal velocity of the object is same as of plane.

Since friction is ignored, horizontal velocity remains unchanged.

Vx = 350m/s

but the Vy increases due to gravitational acceleration

V^2 - U^2 = 2as

V: FINAL VELOCITY

U: INITIAL VELOCITY

U = 0 m/s

`v=√(2as)= √(2(9.81)(22000)) = 656.99 m/s`