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 +14.2 m/s and measures a time of 15.9 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

Answer:

a= -1.8m/s²

Explanation:

The equation for position x for a constant acceleration a, time t, initial velocity v₀ and initial position x₀ is given by:

$x=\frac{1}{2}at^2+v_0t+x_0$

For the case that x = x₀, the equation becomes:

$0=\frac{1}{2}at^2 +v_0t$

Solving for a:

$a=-\frac{2v_0}{t}$