1. What is the acceleration on the Earth?
Remember the equation for an object that starts at rest is this:
d = 1/2at2
( this is an experiment for droping a ball from a 6 meter hight and seeing how long it takes to get to the floor from 6 meters, it took 1.9 seconds to reach the floor, so how do i find the acceleration of the ball?)
I get messed up on this question.
d = 6 m t = 1.9s a = g g = -9.81m/s^2
d = ½ ( -9.81m/s^2)(1.9s)^2 = -17.70
now what do I do from here how do I find the acceleration?
Go back and look at what you wrote. You are on the right path. You already know the value of d. It is 6m. You should not have put the value in for a, the acceleration. That is what you are trying to find.
6m = (1/2) * (a)* (1.9s)^2
Solve for a
3.32 m/s2
To find the acceleration of the ball, you need to rearrange the equation:
d = 1/2at^2
First, multiply both sides of the equation by 2 to eliminate the fraction:
2d = at^2
Next, divide both sides of the equation by t^2 to isolate the acceleration:
a = (2d) / t^2
Now, plug in the given values into the equation:
d = 6 m (distance)
t = 1.9 s (time)
a = (2 * 6 m) / (1.9 s)^2
Simplifying the equation:
a = 12 m / (3.61 s^2)
Finally, calculate the acceleration:
a ≈ 3.32 m/s^2
Therefore, the approximate acceleration of the ball is 3.32 m/s^2.