Answer:
Approximately
(assuming that the projectile was launched at angle of
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing
is the same as the altitude
at which this projectile was launched:
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is
(upwards,) the vertical velocity right before landing would be
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be
. In other words,
.
Hence, the time it takes to achieve a (vertical) velocity change of
would be:
.
Hence, this projectile would be in the air for approximately
.
Answer:
1.28 x 10^4 N
Explanation:
m = 1500 kg, h = 450 km, radius of earth, R = 6400 km
Let the acceleration due to gravity at this height is g'
g' / g = {R / (R + h)}^2
g' / g = {6400 / (6850)}^2
g' = 8.55 m/s^2
The force between the spacecraft and teh earth is teh weight of teh spacecraft
W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N