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An object is thrown upward from the top of 80 ft tower.The height h of the object after t seconds is represented by the quadratic equation h=-16t^2+64t+80.After how many seconds will the object hit the ground?

Verified Answer :

bartdrinksmalk

\(h=-16t^{2}+64t+80\)

We set \(h=0\):
\(-16t^{2}+64t+80=0\)
\(\Rightarrow t^{2}-4t-5=0\)
\(\Rightarrow (t+1)(t-5)=0\)
\(\Rightarrow t=-1\) or \(t=5\)
We use \(t=5\) because a negative number wouldn't make sense

So the object hits the ground after 5 seconds.

Answer:The object will hit the ground after 5 seconds


Step-by-step explanation:

Given height of tower =80 ft

h=-16\(t^{2}\)+64t+80

When the object will hit the ground its height=0

0=-16\(t^{2}\)+64t+80

Multiplying both sides by -1

16t²-64t-80=0

Dividing by 16 both sides

t²-4t-5=0

(t-5)(t+1)=0

t=5 or -1

So t=5 since time cannot be negative

The object will fall after 5 seconds



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