An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 19.0 m if her initial speed is 3.80 m/s. What is the free-fall acceleration on the planet?

Answer 1

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 = 0.96m/s2.

Answer 2

.56