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+2, where
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Question 1002255: 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+2, where h is in feet. If the object is propelled from a height of 4 feet with an initial velocity of 96 feet per second, its height is given by the equation h=16t^2+96t+4.
After how many seconds is the height 112 feet?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The coefficient of t^2 is gravity so it is negative. (Upward force is positive)
:
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 a height of 4 feet with an initial velocity of 96 feet per second, its height is given by the equation h=-16t^2+96t+4.
After how many seconds is the height 112 feet?
:
Write it
-16t^2 + 96t + 4 = 112
-16t^2 + 96t + 4 - 112 = 0
-16t^2 + 96t - 108 = 0
Simplify, divide eq by 4
-4t^2 + 24t - 27 = 0
use the quadratic formula, a=-4; b=24; c=-27,
I obtained two valid solutions
t = 1.5 sec, at 112 ft on the way up
and
t = 4.5 sec, on the way down
:
Looks like this graphically, green line is 112 ft
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