Final answer:
It takes 2.25 seconds for a ball to hit the ground when dropped from a height of 81 feet, assuming there is no initial velocity and only the force of gravity is acting on it.
Step-by-step explanation:
To determine after how many seconds the ball hits the ground from a height of 81 feet, we can use the equation of motion for free-falling objects: h(t) = -16t² + vt + s, where h(t) is the height as a function of time t, -16 is half the acceleration due to gravity in ft/s² (since we are working with feet and not meters), v is the initial velocity, and s is the initial height.
Since the ball is dropped and not thrown, the initial velocity (v) is 0. This simplifies the equation to: h(t) = -16t² + s. For h(t) to be 0 (when the ball hits the ground), we need to find the positive root of the equation:
0 = -16t² + s
We set s to 81 feet, since that is the initial height, and solve the quadratic equation:
0 = -16t² + 81
When we solve this, we get:
t² = 81 / 16
t = √(81 / 16)
t = √5.0625
t = 2.25
Therefore, it takes 2.25 seconds for the ball to hit the ground.