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
