What is the free fall acceleration on the planet?

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SUMMARY

The free fall acceleration on the planet is calculated to be 0.6 m/s² based on the astronaut's maximum horizontal jump distance of 15 m with an initial speed of 3 m/s. The formula used is R = Vi²sin2α/g, where the optimal launch angle for maximum range is determined to be 45 degrees. This angle maximizes the range due to the sine function reaching its peak value at 90 degrees. The calculations confirm that the gravitational acceleration on this planet is significantly lower than Earth's.

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Homework Statement



An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 15 m if her initial speed is 3 m/s. What is the free fall acceleration on the planet?

Homework Equations



R=Vi^2sin2α/g

The Attempt at a Solution



15=9sin2α/g g=9sin2α/15
 
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Solution:
To have maximum range for a given initial velocity, her launch angle must be q0 = 45o. Her range then is R = (v02sin2q 0)/g' = v02sin90o/g' = v02/g'. We have g' = v02/R = 0.6m/s2.

Source: http://electron9.phys.utk.edu/phys135d/modules/m3/Projectile%20motion.htm

Scroll down to almost at the bottom of the page.
 
Last edited by a moderator:
Thanks for clarifying things up. I never knew that launching a projectile at 45 degrees would result in the greatest maximum range.
 

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