SOLUTION: 22. The equation h=-16t^2+112t give the height of an arrow, shot upward from the ground with an initial velocity of 112ft/s, where t is the time after the arrow leaves the ground.

Question 61488: 22. The equation h=-16t^2+112t give the height of an arrow, shot upward from the ground with an initial velocity of 112ft/s, where t is the time after the arrow leaves the ground. Find the time it takes for the arrow to reach a a height of 180 ft.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
22. The equation h=-16t^2+112t give the height of an arrow, shot upward from the ground with an initial velocity of 112ft/s, where t is the time after the arrow leaves the ground. Find the time it takes for the arrow to reach a a height of 180 ft.
180 = -16T^2+112T
DIVIDING WITH 4
4T^2-28T+45=0
4T^2-10T-18T+45=0
2T(2T-5)-9(2T-5)=0
(2T-5)(2T-9)=0
T=2.5 OR 4.5.....
THIS MEANS THAT IT WILL BE AT 180 FEET FROM GROUND AFTER IT WAS SHOT FROM THE GROUND
1. AFTER 2.5 SECS WHILE IT IS GOING UP.
2.THEN AFTER REACHIBG THE MAXIMUM HEIGHT , IT WILL FALL BACK TO THE GROUND
AND SO AFTER 4.5 SECS REFERS TO ITS DOWN WARD FALL

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