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 +14 m/s and measures a time of 17.8 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

local acceleration due to gravity is -1.6 m/s The rock will have a constant deceleration due to the force of gravity and will travel for half the time until it reaches its peak altitude and then fall for the remaining time until it impacts the astronaut's hand. So first calculate how long it takes for the rock to reach its peak altitude by dividing the total time by 2. 17.8 s / 2 = 8.9 s Now the rock's velocity changes from +14 m/s to 0 m/s over a time period of 8.9 s. That gives us the following equation 14 m/s + 8.9x s = 0 Solve for x by first subtracting 14 m/s from both sides 8.9x s = -14 m/s Now divide both sides by 8.9 s x = -1.573 m/s^2 Since we only have 2 significant digits, round the result to 2 significant digits. x = -1.6 m/s