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We throw an object upward from the top of a 912 feet tall building. The vertical position h of the object, (measured in feet) t seconds after we threw it is h=_16t^2+256t+912 How long does it take for the object to hit the ground?

User PhilG
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Final answer:

To determine the time it takes for an object thrown from a 912 feet tall building to hit the ground, we solve the quadratic equation h = -16t^2 + 256t + 912 by setting h to zero. Using the quadratic formula, we obtain the time t as 19 seconds, which is the time taken for the object to reach the ground.

Step-by-step explanation:

To find out how long it takes for the object to hit the ground, we need to solve the given quadratic equation for time, where h (the height) is equal to zero since that's the height when the object hits the ground. The equation given is h = -16t^2 + 256t + 912, and we set h = 0 for the impact:

0 = -16t^2 + 256t + 912

This quadratic equation can be solved using the quadratic formula t = (-b ± √(b^2 - 4ac)) / (2a), where a = -16, b = 256, and c = 912. By substituting these values into the formula, we will get two solutions for t, and the positive value will represent the time it takes for the object to reach the ground.

Applying the quadratic formula, we have:

t = (-256 ± √(256^2 - 4 * (-16) * 912)) / (2 * -16)

t = (-256 ± √(65536 + 58368)) / (-32)

t = (-256 ± √(123904)) / (-32)

t = (-256 ± 352) / (-32)

We get two possible times, but we discard the negative one since time cannot be negative for this scenario. Therefore, the positive root is the correct time to hit the ground:

t = (256 + 352) / 32

t = 608 / 32

t = 19 seconds

It takes 19 seconds for the object to hit the ground after being thrown upwards from the top of the 912 feet tall building.

User Peter Catalin
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