SOLUTION: an object is thrown upward from the top of a 48-foot building with an initial velocity of 32 feet per second. the height h of the object after t seconds is given by the quadratic e

Question 949108: an object is thrown upward from the top of a 48-foot building with an initial velocity of 32 feet per second. the height h of the object after t seconds is given by the quadratic equation h=-16t^2+32t+48. when will the object hit the ground?
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
we are given
h=-16t^2+32t+48
when object hits the ground, h = 0, therefore
0 = -16t^2+32t+48
divide both sides of = by 16
-t^2 +2t + 3 = 0
factor equation
(-t+3)(t+1) = 0
then we have t = 3 and t = -1
we want the positive value for t
the object will hit the ground in 3 seconds

RELATED QUESTIONS
An Object is thrown upward from the top of a 48-foot building with an initial velocity of (answered by Alan3354,Boreal)

An object is thrown upward from the top of a 160-foot building with an initial velocity... (answered by Alan3354)

An object is thrown upward from the top of a 160 foot building with an initial velocity... (answered by mananth)

an object is thrown upward from the top of a 64 foot building with an initial velocity of (answered by ikleyn)

7) The path of a falling object is given by the function where represents the initial... (answered by stanbon)

An object is thrown upward from the top of a 260 foot tall building with an initial... (answered by solver91311)

An object is thrown upward from the top of a 96 foot building with an initial velocity of (answered by josmiceli)

An object is thrown upward from the top of a 160-foot building with an initial... (answered by ankor@dixie-net.com)

An object is thrown upward from the top of a 6464-foot building with an initial... (answered by Alan3354)