Question

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?

Answer 1

To determine how long it takes for the object to hit the ground, we need to find the time at which the height (h) becomes zero.

Given the equation for the vertical position of the object: h = -16t^2 + 256t + 912

Setting h to zero and solving for t: 0 = -16t^2 + 256t + 912

This is a quadratic equation, which we can solve by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula in this case:

t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, the coefficients are: a = -16 b = 256 c = 912

Plugging in these values into the quadratic formula, we have: t = (-256 ± √(256^2 - 4(-16)(912))) / (2(-16))

Simplifying further: t = (-256 ± √(65536 + 58368)) / (-32) t = (-256 ± √(123904)) / (-32) t = (-256 ± 352) / (-32)

Now we have two possible solutions for t:

1. t = (-256 + 352) / (-32)
2. t = (-256 - 352) / (-32)

Calculating each solution:

1. t = (96) / (-32) = -3
2. t = (-608) / (-32) = 19

Since we are looking for the time it takes for the object to hit the ground, we discard the negative value (-3) as it is not meaningful in this context.

Therefore, it takes approximately 19 seconds for the object to hit the ground.