As the ball rises to its maximum height, its velocity decreases from its initial velocity to 0 m/s at the rate of 9.8 m/s^2. As the ball falls from its maximum, its velocity increases from 0 m/s to its final velocity at the rate of 9.8 m/s^2. Since height of the timer is constant, the time up is equal to the time down. And, the magnitudes of the initial and final velocities are equal in magnitude.
u = g(t1 + t2)/2
Since t1 and t2 are equal, (t1 + t2)/2 is t
u = g * t
u – g * t = 0
This is what my first sentence says.
[2√(u² - 2gh)]/g.
This equation comes from the following equation.
vf^2 = vi^2 + 2 * a * d
vf is 0, a is g (-9.8)
d is the maximum height.
u² - 2gh is 0. Like I said, the two times are equal.
Hence, find a formula for h in terms of t1, t2 and g.
d = vi * t – ½ * g * t^2