SOLUTION: if an object is propelled upward from a height of s feet at an initial velocity of v feet per second, then its height h after t seconds is given by the equation h=-16t^2+vt+s, wher
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Question 784982: if an object is propelled upward from a height of s feet at an initial velocity of v feet per second, then its height h after t seconds is given by the equation h=-16t^2+vt+s, where h is in feet. If the object is propelled from height of 4 feet with an initial velocity of 96 feet per second, its height h is given by the equation h= -16t^2+64t+8. After how many seconds is the height 112 feet?
You can put this solution on YOUR website! if an object is propelled upward from a height of s feet at an initial velocity of v feet per second, then its height h after t seconds is given by the equation h=-16t^2+vt+s, where h is in feet.
If the object is propelled from height of 4 feet with an initial velocity of 96 feet per second, its height h is given by the equation h= -16t^2+64t+8. After how many seconds is the height 112 feet?
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From the given information we can write the equation:
-16t^2 +96t + 4 = 112
-16t^2 + 96t + 4 - 112 = 0
-16t^2 + 96t - 108 = 0
simplify, change the signs, divide by -4
4t^2 - 24t + 27 = 0
You can use the qudratic formula but this will factor to
(2t-3)(2t-9) = 0
Two solutions
t = 3/2
t = 1.5 seconds at 112 ft on the way up
and
t = 9/2
t = 4.5 seconds at 112 ft on the way back down
Graphically, (green line is 112 ft)
