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an object is thrown upward from the top of a 160 foots building with an initial velocity of 48 feet per second .solve the equation -16^2 + 48t + 160=0 find the time(t) in seconds at which the object hits the ground.

User Vekah
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1 Answer

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The given equation represents the distance travelled by the object at time t.

The given equation is expressed as

-16t^2 + 48t + 160

At the point where it hits the ground, the distance woule be 0. Thus, we would so;ve the equation,

-16t^2 + 48t + 160=0

We would divide through by - 16. We have

t^2 - 3t - 10 = 0

We would find two terms such that their sum or difference is - 3t and their product is - 10t^2. They are 2t and - 5t. We have

t^2 + 2t - 5t - 10 = 0

By factorising, we have

t(t + 2) - 5(t + 2) = 0

(t + 2)(t - 5) = 0

t + 2 = 0 or t - 5 = 0

t = - 2 or t = 5

Since the time cannot be negative, the correct answer is

time = 5 seconds

User Animeta
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