An object is launched directly in the air at a speed of 32 feet per second from a platform located 8 feet in the air. The motion of the object can be modeled using the function f (t) = -16t2+32t+8, where t is the time in seconds and f (t) is the height of the object. How long, in seconds, will the object be in the air before hitting the ground? Round your answer to the nearest hundredth of a second. Do not include units in your answer.

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Mathematics

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30 November, 23:08

An object is launched directly in the air at a speed of 32 feet per second from a platform located 8 feet in the air. The motion of the object can be modeled using the function f (t) = -16t2+32t+8, where t is the time in seconds and f (t) is the height of the object. How long, in seconds, will the object be in the air before hitting the ground? Round your answer to the nearest hundredth of a second. Do not include units in your answer.

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Reina Wolfe · Answer 1

2.225

Step-by-step explanation:

To find how long the object will be in the air, we just need to find the time when the object reaches the ground, that is, f (t) = 0

So, we have that:

0 = - 16t2+32t+8

2t2-4t-1 = 0

Using Bhaskara's formula:

Delta = b2 - 4ac = 16 + 8 = 24

sqrt (Delta) = 4.899

t = (-b + sqrt (Delta)) / 2a

t = (4 + 4.899) / 4

t = 2.2248

Rounding to nearest hundredth, we have t = 2.225