How High and Far Does a Spring-Propelled Space Shuttle Travel?

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The discussion focuses on calculating the maximum height and horizontal distance traveled by a spring-propelled space shuttle with a mass of 3500 kg and a spring constant of 1600 N/m. The spring compresses 450 m on a frictionless ramp inclined at 45 degrees, with a height of 30 m. The total energy conservation equation is applied, leading to a calculated velocity of approximately 303.2 m/s. However, the resulting height of 2375 m appears unrealistic, indicating potential errors in the calculations. The user expresses uncertainty about determining the time of flight for calculating horizontal distance, suggesting a need for further clarification on the problem-solving approach.
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Homework Statement


The Lower Machiavellian Space Shuttle, having a total mass of 3500kg, is about to make its' debut launch. Rather then using a rocket engine to launch the shuttle, the Machiavellianans' use a large compressed spring, having a spring constant of 1600N/m. If the spring can be com[pressed by a distance of 450m and if the launching ramp is frictionless, calculate
a)the max height the space shuttle obtains, and
b)the horizontal distance the shuttle lands, as measured from the base of the ramp

*also the ramp is 45 degrees and has a height of 30m

Homework Equations


Ee=1/2kx^2
Ek=1/2mv^2
Eg=mgh

The Attempt at a Solution


Etot=Etot'
Ee=Ek+Eg
1/2kx^2=1/2mv^2+mgh
.5(1600)(450^2)=.5(3500)(v^2)+3500(9.8)(30)
162,000,000=1750v^2+1,029,000
162,000,000-1,029,000=1750v^2
160,971,000/1750=v^2
v^2=91983.4
v=303.2

from there I did sin(45)*303.2 which gave me a horz and vert speed of 214.4

vert:
1/2mv^2=mgh *im assuming 30m is ground level*
.5(214.4^2)(3500)=3500(9.8)(h)
80442880=34300h
h=80442880/34300
h=2345.27+30=2375
*that height seems very unrealistic so I'm assuming I did something wrong
 
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here*horz:v=d/t303.2=d/td=303.2tI'm not sure how to calculate t in this case so I'm assuming I'm wrong here too
 
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