Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 3.00 m above the ground and measure that it hits the ground 0.814 s later What is the acceleration of gravity near the surface of this planet?

Answer: The gravitational acceleration near the surface of this planet is 9.056 m/s^2

Explanation:

When you drop an object, the only acceleration acting on the object is the gravitational acceleration, so we can writhe the acceleration of the wrench as

a (t) = - g

where the minus sign is because the wrench will move downwards.

for the velocity of the object, we integrate over time and get:

v (t) = - g*t + v0

where v0 is the initial velocity, in this case is 0, because the wrench is droped.

integrating again, we will get the position of the wrench:

p (t) = - (g/2) * t^2 + p0

where p0 is the initial position, in this case is 3m above the ground.

p (t) = - (g/2) * t^2 + 3m

now, afther 0.814seconds the wrench hits the ground, so we have:

p (0.814 s) = 0 = - (g/2) * (0.814s) ^2 + 3m

now we need to solve this for g.

- (g/2) * (0.814s) ^2 + 3m = 0

(g/2) * (0.814s) ^2 = 3m

(g/2) = 3m / * (0.814s) ^2 = 4.528m/s^2

g = 2*4.528m/s^2 = 9.056 m/s^2