Question

An object is thrown upward from the top of 96​-foot building with an initial velocity of 80 feet per second. The height h of the object after t seconds is given by the quadratic equation h=16t^2+80t+96. When will the object hit the​ ground?

Answer 1

The function of the object height is an illustration of a projectile motion

The object will never hit the ground

How to determine when the object hits the ground?

The function is given as:

h=16t^2+80t+96.

When the object hits the ground, h = 0

So, we have:

16t^2+80t+96 = 0

Divide through by 16

t^2+5t+6 = 0

Expand

t^2 + 3t + 2t + 6 = 0

Factorize

t(t + 3) + 2(t + 3) = 0

Factor out t + 3

(t + 2)(t + 3) = 0

Solve for t

t = -2 or t = -3

The time (t) cannot be negative.

Hence, the object will never hit the ground

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