SOLUTION: An object projected from height of 48 ft. with an initial velocity of 32 ft./sec after t seconds has a height = -16t^2 + 32t +48.
A) After how many seconds is the height 64 ft?
B
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-> SOLUTION: An object projected from height of 48 ft. with an initial velocity of 32 ft./sec after t seconds has a height = -16t^2 + 32t +48.
A) After how many seconds is the height 64 ft?
B
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Question 907730: An object projected from height of 48 ft. with an initial velocity of 32 ft./sec after t seconds has a height = -16t^2 + 32t +48.
A) After how many seconds is the height 64 ft?
B) After how many seconds does the object hit the ground?
C) How high is the object after 1/2 second?
I'm not entirely confident with my answers but here is what I got:
A) 1 second
B) 3 seconds
C) 60 feet
I just need help to see if I understood this correctly. I was having a very difficult time with this problem. Answer by KMST(5328) (Show Source):
You can put this solution on YOUR website! THE REASONING:
A) --->
You solved that equation correctly and found .
B) --->
You solved that equation correctly and found .
C) ---> ,
and you did the calculation correctly.
THE QUADRATIC EQUATIONS:
There are many ways to solve. This is my way.
(Here I would say that I am dividing both sides of the equal sign by -16,
but I think of it as two steps)
Dividing both sides of the equation by 16, we get , and changing the signs (multiplying both sides times -1), we get
Adding 1 to both sides, I complete the square: --->--->
