An object is thrown upward from the top of a 160-foot building with an initial v

Question

An object is thrown upward from the top of a 160-foot building with an initial velocity of 48 feet per second. The height h of the object after t seconds is given by the quadratic equation h= -16t^2 + 48 + 160. When will the object hit the ground? An object is thrown upward from the top of a 160-foot building with an initial velocity of 48 feet per second. The height h of the object after t seconds is given by the quadratic equation h= -16t^2 + 48 + 160. When will the object hit the ground? of a 160-foot building with an initial velocity of 48 feet per second. The height h of the object after t seconds is given by the quadratic equation h= -16t^2 + 48 + 160. When will the object hit the ground?

Explanation / Answer

Object will hit the ground when h=0.

-16t2+48t+160=0

so,

-t2+3t+10=0

so,

t2-3t-10=0

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

t=5,-2

But t=-2 is not possible.

SO t=5s is correct.

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