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A person is about to drop a ball off'of a balcony. The balcony is 81 feet above the ground. After how many seconds does the ball hit the ground? Recall: h(t) =-161? + v+ s

User Verona
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1 Answer

5 votes

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.

User Candrews
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