Answer to Question #244648 in Physics for Have a good day

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 + 15.7 m/s and measures a time of 18.7 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet? (positive = up, negative = down)

Answer 1

$v_y=v_{0y}-g_p\dfrac{t_{tot}}{2},$ $0=v_{0y}-g_p\dfrac{t_{tot}}{2},$ $g_p=\dfrac{2v_{0y}}{t_{tot}}=\dfrac{2\cdot15.7\ \dfrac{m}{s}}{18.7\ s}=1.68\ \dfrac{m}{s^2}.$