To determine the time it takes for the ball to reach its maximum height, we can use the equation of motion that relates displacement (d), initial velocity (v₀), time (t), and acceleration (a):
d = v₀t + (1/2)at²
In this case, the ball is thrown upward, so the initial velocity (v₀) is 24 ft/s, the acceleration (a) due to gravity is approximately -32.2 ft/s² (negative because it acts in the opposite direction of the upward motion), and the displacement (d) is the change in height, which is +50 ft (taking upward as the positive direction). The goal is to find the time (t).
Plugging in these values into the equation:
50 ft = (24 ft/s)t + (1/2)(-32.2 ft/s²)t²
Simplifying the equation:
50 ft = 24 ft/s * t - 16.1 ft/s² * t²
Rearranging the equation:
16.1 ft/s² * t² - 24 ft/s * t + 50 ft = 0
This is a quadratic equation in terms of time (t). Solving this equation will yield the time it takes for the ball to reach its maximum height. Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
For this equation, a = 16.1 ft/s², b = -24 ft/s, and c = 50 ft. Plugging these values into the quadratic formula:
t = (-(-24 ft/s) ± √((-24 ft/s)² - 4 * 16.1 ft/s² * 50 ft)) / (2 * 16.1 ft/s²)
Simplifying further:
t = (24 ft/s ± √(576 ft²/s² - 3228.4 ft²/s²)) / 32.2 ft/s²
t = (24 ft/s ± √(-2652.4 ft²/s²)) / 32.2 ft/s²
The value inside the square root is negative, which means there are no real solutions for time in this case. It indicates that the ball will not reach a maximum height and will fall back to the ground before that happens. Therefore, the time to reach the maximum height cannot be determined.
As for the distance above the ground, since the ball does not reach its maximum height, it will remain at a height of 50 ft throughout its motion.