contestada

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?

Respuesta :

Answer:

Vx = 350m/s

Vy  = 656.99 m/s

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

[tex]v=\sqrt{2as}= \sqrt{2(9.81)(22000)} = 656.99 m/s[/tex]

Otras preguntas