To solve this problem, we can use the equations of motion for an object in free fall.
Let's denote:
- h as the height of the tower (in meters)
- t as the total time the ball remains in the air (in seconds)
From the equations of motion, we know that the height (s) of an object in free fall can be expressed as:
s = ut + (1/2) * g * t^2
where:
- u is the initial velocity (which is zero when the ball is dropped)
- g is the acceleration due to gravity (which is approximately 10 m/s²)
For the first part of the motion, from the top of the tower to a height of h/2, the equation becomes:
(h/2) = (1/2) * g * t^2
Simplifying:
t^2 = h / g
t = sqrt(h / g)
Now, for the second part of the motion, from a height of h/2 to the ground, the equation becomes:
(h/2) = (1/2) * g * (t - 1)^2
Simplifying and expanding:
(h/2) = (1/2) * g * (t^2 - 2t + 1)
h = g * (t^2 - 2t + 1)
Rearranging the equation:
t^2 - 2t + 1 = h / g
t^2 - 2t + (1 - h / g) = 0
This is a quadratic equation in terms of t. We can solve it using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case: a = 1 b = -2 c = 1 - h / g
Substituting the values into the quadratic formula:
t = (-(-2) ± sqrt((-2)^2 - 4 * 1 * (1 - h / g))) / (2 * 1)
t = (2 ± sqrt(4 - 4(1 - h / g))) / 2
t = (2 ± sqrt(4 - 4 + 4h / g)) / 2
t = (2 ± sqrt(4h / g)) / 2
t = 1 ± sqrt(h / g)
Since time cannot be negative, we take the positive solution:
t = 1 + sqrt(h / g)
Therefore, the ball remains in the air for a total time of 1 plus the square root of h divided by g seconds.