A projectile is shot from the surface of earth by means of a very powerful cannon. if the projectile reaches a height of 100.0 km above earth's surface, what was the speed of the projectile when it left the cannon?

Respuesta :

The relevant formula we can use in this case is:

v^2 = v0^2 + 2 a d

where,

v = final velocity = 0

v0 = initial velocity = ?

a = acceleration = -9.8 m/s^2 (negative since opposite)

d = 100 km = 100,000 m

 

0 = v0^2 + 2 (-9.8) (100,000)

v0 = 1400 m/s