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2 h(t) = – 16 t + vt + s  v  initial upward velocity (in ft. per sec.)  s  initial height (in feet)  t  time the object is in motion (in seconds)

0 = – 16 t + 48 t h(t) = – 16 t + vt + s 2  v  initial upward velocity (in ft. per sec.)  s  initial height (in feet)  t  time the object is in motion (in seconds) You are standing on a cliff that is 154 feet above the Willamette River. You throw a rock with an initial velocity of 48 feet per second. Your hand was 6 feet above the cliff when the rock was released. How long did it take for the rock to hit the water? Example: 1 st identify v and s v = 48 feet per second S =  160 feet Note: h = 0, the height of the rock at impact 2 nd Substitute into the formula h(t) = – 16 t + vt + s 2 2

Example Continued 1 st identify v and s v = 48 feet per second S =  160 feet Note: h = 0, the height of the rock at impact 2 nd Substitute into the formula h(t) = – 16 t + vt + s rd Factor 0 = – 16 t + 48 t Factor out greatest common monomial first. 2 Factor remaining trinomial 0 = (– 16)( t – 5)( t + 2) Zero product property t – 5 = 0 and t + 2 = 0 t = 5 and t = – 2 Final answer It will take 5 Seconds for the rock to hit the water. Time cannot be negative, so t = − 2 is invalid

0 = – 16t + 24t Example: 1 st identify v and s v = 24 feet per second S =  1840 feet 2 nd Substitute 2 3 rd Factor 0 = – 16t + 24t Factor out greatest common monomial first 0 = (– 8)(2t – 3 t – 230 ) 2 Factor remaining trinomial 0 = (– 8)(2t – 23)( t + 10) Zero product property 2t – 23 = 0 and t + 10 = 0 2t = 23 and t = – 10 v  initial upward velocity (in ft. per sec.) s  initial height (in feet) t  time the object is in motion (in seconds) h(t) = – 16t + vt + s 2 My son threw my cell phone from the rim of the Grand Canyon. He threw it with an initial velocity of 24 feet per second. The phone hit 1836 feet below our feet. His hand was 4 feet above the ground when the phone was released. How long did it take for my cell phone to smash on the ground. t = 23 2 or 11.5 Final answer It will take 11.5 Seconds for the phone to hit the ground. Time cannot be negative, so t = −10 is invalid

± 9 = t 2 nd Substitute Example: My daughter was in Borneo at the base of a fig tree when an orangutan dropped a fig from the top of the canopy. If the orangutan dropped the fig from 144 feet above her feet, how long did it take to hit the ground in front of her? 0 = – 16t st identify v and s v = 0 feet per second S = 144 feet 2 3 rd Solve using square roots 0 = – 16t Since b = 0, isolate t and use square roots to solve. h(t) = – 16t + s 2 Final answer: It will take 3 Seconds for the fig to hit the ground. Time cannot be negative, so t = − 3 is invalid s  initial height (in feet) t  time the object is in motion (in seconds) Formula with no initial velocity 2 – 144 = – 16t 2 Subtract 144 Divide by = t 2 Square root 2 ± 3 = t t = 3 Solution

2 h(t) = – 16 t + vt + s 2  v  initial upward velocity (in ft. per sec.)  s  initial height (in feet)  t  time the object is in motion (in seconds) h(t) = – 16 t + s 2  s  initial height (in feet)  t  time the object is in motion (in seconds) Summary: