Question

An astronaut on a distant planet wants to determine its acceleration due to gravity. the astronaut throws a rock straight up with a velocity of +16 m/s and measures a time of 21.2 s before the rock returns to his hand. what is the acceleration (magnitude and direction) due to gravity on this planet? (indicate direction by the sign of the acceleration.)

Answer 1

Define
u = 16 m/s, the vertical launch velocity
g = acceleration due to gravity, measured positive downward
s = vertical distance traveled
t = 21.2 s, total time of travel.

The vertical motion obeys the equation
s = ut - (1/2)gt²

When the rock is at ground level, s = 0.
Therefore
(16 m/s)(21.2 s) - 0.5*(g m/s²)*(21.2 s)² = 0
339.2 - 224.72g = 0
g = 1.5094 m/s²

Answer:
The acceleration due to gravity is 1.509 m/s² measured positive downward.