Asked by Anonymous
Find the minimum initial speed of a projectile in order for it to reach a height of 2039 km above the surface of the earth.
Answers
Answered by
Matt
2039 km = 2039000
NOTE: Projectile motion signifies an x and y displacement, but let us assume this pertains to the freefall laws of motion.
Top of the trajectory: Vf = 0
Vf^2 = Vo^2 + 2gy
0 = Vo^2 + 2(-9.81 m/s^2)(2039000 m)
V0^2 = 4.00052*10^7 m^2/s^2
Vo = 6324.97 m/s
NOTE: Projectile motion signifies an x and y displacement, but let us assume this pertains to the freefall laws of motion.
Top of the trajectory: Vf = 0
Vf^2 = Vo^2 + 2gy
0 = Vo^2 + 2(-9.81 m/s^2)(2039000 m)
V0^2 = 4.00052*10^7 m^2/s^2
Vo = 6324.97 m/s
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