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 8

equals -16t 2nd power + 48t + 160 when will the object hit the ground

s thrown upward Nom the top of a 160 foot building with an I equation h -16 40 160 When wil the object hit the ground? ritial velocity of 48 feot per second The heighth of the object after t seconds is given by t 0 sec ec ec 2 sec ect your answer Eamed amaery scove on Secton 8 3 Homewn

Explanation / Answer

h(t)= -16t² +48t +160

when object hit the ground, height will be 0,

-16t² +48t +160 = 0

Divide both sides by -16

t² -3t -10 = 0

Factor it as

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

t= 5

Hence after 5 seconds

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