A rocket is fired straight up through the atmosphere from the South Pole, burning out at an altitude of 239 km when traveling at 5.00 km/s.
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At Ux = 0 and Uy = U = 5 kps when reaching H = 239 km above the S pole, it has no X velocity component. So it will fall straight back down and destroy the launch site and everyone around it, including the penguins.
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Maybe that's escape velocity and it just keeps on truckin'. Let's test that WAG.
From 1/2 mU^2 = mGM/(R + h)^2 * (R + h); we have U = sqrt(2GM/(R + h)) = sqrt(2*6.67E-11*5.96E24/((6340 + 239)*1000)) = 10993.12336 = 10993 mps. So the 5000 mps is not escape velocity.
It just falls back down from where it started (discounting wind and such) and ruins the launch pad and everyone's day.
So the max distance is the max height above the launch pad. And that's:
Height at any time t = y(t) = h + Uy t - 1/2 g t^2
Distance at any time t = x(t) = Ux t
g = 9.810 m/s^2
Launch height above ground H = 239000.000 meters
Impact (target) elevation y(T) = 0.000 meters
h = H - y(T) = 239000.000 meters
Launch speed U = 5000.000 mps 18000.000 kph
Launch angle theta = 90.000 degrees
Uy = U sin(theta) = 5000.000 mps 18000.000 kph
Ux = U cos(theta) = 0.000 mps 0.000 kph
For total flight time T, solve the quadratic: 0 = 239,000.000 + 5,000.000 T - 4.905 T^2
Quadratic coefficients: A = 4.905 B = -5000.000 C = -239000.000
Total Flight Time sec
T = 1065.115
T' = -45.747
Max Range to Impact meters
x(T) = Ux T = 0.000 * 1,065.115 = 0.000
Range at Max Height meters
x(tmax) = 0.000
Max Height above impact elevation meters
y(max) = 1513209.990 <=================== ANS. [See source.]
Max Height above h meters
y(max) - h = 1274209.990
Time to reach max height sec
tmax = 509.684
Vy at impact mps
Vy = Uy - gT = 5,000.000 - 9.810 * 1,065.115 = -5448.778
Vx at impact mps
Vx = Ux = 0.000
Total impact speed mps
V = sqrt(Vy^2 + Vx^2) = 5448.778
Angle of impact re horizontal degrees
psi = ATAN(Vy/Vx) = ATAN(-5,448.778/0.000) = -90.000
OR
Maybe that's escape velocity and it just keeps on truckin'. Let's test that WAG.
From 1/2 mU^2 = mGM/(R + h)^2 * (R + h); we have U = sqrt(2GM/(R + h)) = sqrt(2*6.67E-11*5.96E24/((6340 + 239)*1000)) = 10993.12336 = 10993 mps. So the 5000 mps is not escape velocity.
It just falls back down from where it started (discounting wind and such) and ruins the launch pad and everyone's day.
So the max distance is the max height above the launch pad. And that's:
Height at any time t = y(t) = h + Uy t - 1/2 g t^2
Distance at any time t = x(t) = Ux t
g = 9.810 m/s^2
Launch height above ground H = 239000.000 meters
Impact (target) elevation y(T) = 0.000 meters
h = H - y(T) = 239000.000 meters
Launch speed U = 5000.000 mps 18000.000 kph
Launch angle theta = 90.000 degrees
Uy = U sin(theta) = 5000.000 mps 18000.000 kph
Ux = U cos(theta) = 0.000 mps 0.000 kph
For total flight time T, solve the quadratic: 0 = 239,000.000 + 5,000.000 T - 4.905 T^2
Quadratic coefficients: A = 4.905 B = -5000.000 C = -239000.000
Total Flight Time sec
T = 1065.115
T' = -45.747
Max Range to Impact meters
x(T) = Ux T = 0.000 * 1,065.115 = 0.000
Range at Max Height meters
x(tmax) = 0.000
Max Height above impact elevation meters
y(max) = 1513209.990 <=================== ANS. [See source.]
Max Height above h meters
y(max) - h = 1274209.990
Time to reach max height sec
tmax = 509.684
Vy at impact mps
Vy = Uy - gT = 5,000.000 - 9.810 * 1,065.115 = -5448.778
Vx at impact mps
Vx = Ux = 0.000
Total impact speed mps
V = sqrt(Vy^2 + Vx^2) = 5448.778
Angle of impact re horizontal degrees
psi = ATAN(Vy/Vx) = ATAN(-5,448.778/0.000) = -90.000
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There are so many answers where, which one exactly is the exact maximum distance?
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I have no idea which one here is the answer.
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