A stone is dropped from the top of a 83.0 m tall building. How far above the ground is the stone 1.60 seconds before it reaches the ground?

first how long before it hits ground?

83 = 4.9 t^2
t = 4.12 s

so it fell for 4.12 - 1.60 = 2.52 seconds
so how far did it fall in 2.52 seconds?
d = 4.9*(2.52)^2 = 31 meters
so it is
83 - 31 = 52 meters up

To find the distance above the ground that the stone is 1.60 seconds before it reaches the ground, we can use the formula for the height of an object in free fall:

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

where:
- h is the height of the object
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time elapsed

In this case, the stone is dropped from rest, so its initial velocity is 0. Therefore, we can ignore the initial velocity term in the equation.

Plugging in the given values:
g = 9.8 m/s^2
t = 1.60 s

h = (1/2) * 9.8 * (1.60)^2

Now let's solve this equation step by step:

1. Calculate (1.60)^2: (1.60)^2 = 2.56
2. Multiply 2.56 by 9.8: 2.56 * 9.8 = 25.088
3. Divide 25.088 by 2: 25.088 / 2 = 12.544

Therefore, the stone is approximately 12.544 meters above the ground 1.60 seconds before it reaches the ground.