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 13.5 m/s and measures a time of 21.8 s before the rock returns to his hand. What is the acceleration due to gravity on this planet? (also select the direction)

The direction of gravity must obviously down. Otherwise, the rock would never come down.

Of the 21.8 seconds, half the time (10.9 s) is spent going up and the other half is coming back down.

To decelerate from 13.5 m/s to 0 speed in 10.9 seconds, the acceleration of gravity must be
13.5 m/s/10.9 s = ____ m/s^2

LOLOL NO YOU DON'T

To determine the acceleration due to gravity on the distant planet, we can use the kinematic equation for an object thrown upwards:

h = ut - (1/2)gt^2

where:
h is the maximum height reached by the rock,
u is the initial velocity of the rock (13.5 m/s),
g is the acceleration due to gravity on the planet (what we want to find),
and t is the time taken for the rock to reach its maximum height and return to the astronaut's hand (21.8 s).

Since the rock starts and ends at the same height (in the astronaut's hand), the maximum height reached is zero. Plugging in the given values, we get:

0 = (13.5 m/s) * (21.8 s) - (1/2)g * (21.8 s)^2

Simplifying this equation, we get:

(13.5 m/s) * (21.8 s) = (1/2)g * (21.8 s)^2

Rearranging the equation to solve for g, we have:

g = (2 * (13.5 m/s) * (21.8 s)) / (21.8 s)^2

Plugging in the values and performing the calculations, we find that the acceleration due to gravity on the distant planet is approximately 13.5 m/s^2 in the direction opposite to the direction of the throw.