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 +17.4 m/s and measures a time of 17.9 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:

- 1.94 m/s^2

Explanation:

Hello!

We can solve this problem using the equation of the velocity in terms of the time:

$v(t) = v_0 + at$

The velocity when the the rock returns to his hand must be the same but with opposite sign than the initial velocity, that is:

$v(17.9s) = -17.4 m/s$

Therefore:

$-17.4 m/s = 17.4 m/s+a(17.9s)$

Solving for a:

$a = -2\frac{17.4 m/s}{17.9 s}= - 1.94 m/s^2$