A projectile is launched vertically from the surface of the earth with an initia

Question

                             A projectile is launched vertically from the surface of the earth with an initial speed vo= 6.74 km/s. a) What is the magnitude of its acceleration                             (m/s^2) 2330 km above the surface of the earth? b) At what height (km) will it stop and begin to fall back to the surface of the earth?                                                                                                             A projectile is launched vertically from the surface of the earth with an initial speed vo= 6.74 km/s. a) What is the magnitude of its acceleration                             (m/s^2) 2330 km above the surface of the earth? b) At what height (km) will it stop and begin to fall back to the surface of the earth?                                                      

Explanation / Answer

Given

Vo = 6,740 m/s (SI unit from km/s)
g = 9.81 m/s^2

The maximum height that it will achieve is simply

H = Vo^2 / [2g]

H = (6,740 m/s)^2 / [ 2 * (9.81 m/s^2) ]
H = (45427600 m^2/s^2) / [ 19.62 m/s^2 ]
H = 2,315,372 m

H = 2315 kms

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