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 + 15.5 m/s and measures a time of 13.9 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet?

To determine the acceleration due to gravity on the distant planet, you can use the kinematic equation for vertical motion. The equation is given as:

h = v0 * t + (1/2) * g * t^2

Where:
h = vertical displacement (which is zero in this case since the rock returns to the astronaut's hand)
v0 = initial velocity of the rock (which is +15.5 m/s)
t = time taken by the rock to return to the astronaut's hand (which is 13.9 s)
g = acceleration due to gravity on the planet (unknown)

Since the rock returns to the astronaut's hand, the final displacement (h) is zero. Therefore, we can rearrange the equation as follows:

0 = (15.5 m/s * 13.9 s) + (1/2) * g * (13.9 s)^2

Simplifying the equation further:

0 = 215.45 m + 0.5 * g * 193.21 s^2

Now, we can isolate the acceleration due to gravity (g):

-215.45 m = 0.5 * g * 193.21 s^2

Dividing both sides of the equation by 0.5 * 193.21 s^2:

-215.45 m / (0.5 * 193.21 s^2) = g

Calculating the value:

g ≈ -11.16 m/s^2

The negative sign indicates that the acceleration due to gravity is in the opposite direction of the initial velocity. So, the magnitude of the acceleration due to gravity on this planet is approximately 11.16 m/s^2, and its direction is downward.

h=1/2 g t^2 +Vi*t

g=-9.8, h=0 solve for g