Question

An object is thrown upward from the top of a 112 foot building with an initial velocity of 96 feet per second. The height h of the object after t seconds is given by the quadratic equation h=-16t + 96t + 112. When will the object hit the ground ?

Answer 1

h= -16t²+96t+112
=> -16t²+96t+112 =0
divide both sides by -16
=> t²-6t-7=0
=> t²+t -7t-7=0
=> t(t + 1) -7(t + 1)=0
=> (t + 1)(t - 7) = 0
=> t = -1 or 7
=> The object will hit the ground after 7 seconds.
(because -1 seconds makes no sense)

Answer 2

When an object is thrown upward from the top of a 112-foot building with an initial velocity of 96 feet per second, the object will reach the ground after 7 seconds.

Given the height of the object thrown expressed using the equation;

h=-16t² + 96t + 112

The height of the object on  the ground will be zero i.e. h(t) = 0

Substituting the given height into the expression above will give;

0 = -16t² + 96t + 112

-16t² + 96t + 112 = 0

16t² - 96t - 112 = 0

Divide through by 16

t² - 6t - 7 = 0

t² - 7t + t - 7 = 0

t(t - 7) + 1(t - 7) = 0

(t + 1) (t - 7) = 0

t = -1 and 7

Hence when an object is thrown upward from the top of a 112-foot building with an initial velocity of 96 feet per second, the object will reach the ground after 7 seconds.

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