SOLUTION: this is the problem
If an object is propelled upward from the ground level with an initial velocity of 42.9 ft. per sec, it's height h in feet t seconds later is given by the equa
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If an object is propelled upward from the ground level with an initial velocity of 42.9 ft. per sec, it's height h in feet t seconds later is given by the equa
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Question 112724: this is the problem
If an object is propelled upward from the ground level with an initial velocity of 42.9 ft. per sec, it's height h in feet t seconds later is given by the equation h=-16t^2+42.9t. After how many seconds does the object hit the ground? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! f an object is propelled upward from the ground level with an initial velocity of 42.9 ft. per sec, it's height h in feet t seconds later is given by the equation h=-16t^2+42.9t. After how many seconds does the object hit the ground?
:
You know that when the object hits the ground, h = 0
Therefore just put the equation = to 0 and solve for t:
:
-16t^2 + 42.9t = 0
Factor out -t (changes the signs)
-t(16t - 42.9) = 0
Two solutions
t = 0; when the object is propelled upward
and
16t = 42.9
t =
t = 2.68 seconds when the object strikes the ground
