Question

A ball is thrown vertically upward from the top of a building 128 feet tall with an initial velocity of 112 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s = 128 + 112t - 16t² (a) After how many seconds does the ball strike the ground?

Answer 1

To find after how many seconds the ball strikes the ground, the quadratic equation representing the ball's distance from the ground is set to zero, and the quadratic formula is applied to find the time t. The positive root from this calculation will be the answer, as time cannot be negative.

Step-by-step explanation:

The student's question is regarding the time it takes for a ball thrown vertically upward to strike the ground after being launched from the top of a building. The given quadratic equation for the distance s of the ball from the ground s = 128 + 112t - 16t² is used to solve this problem. To find out when the ball strikes the ground, we need to determine when the distance s is equal to 0.

Setting the equation to 0 and solving for t gives us:

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• 0 = 128 + 112t - 16t²

We can rearrange this to form a standard quadratic equation:

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• 16t² - 112t - 128 = 0

To solve for t, we can use the quadratic formula:

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• t = [-(-112) ± √((-112)² - 4(16)(-128))]/(2· 16)

Calculating the discriminant and simplifying the equation will yield two solutions for t. Only the positive value of t makes physical sense because time cannot be negative. This positive value is the time in seconds after which the ball hits the ground. In the context of the question, we are interested in this realistic, physical solution.