SOLUTION: A ball is thrown straight up with an initial velocity of 80 feet per second from the top of a building 23 feet tall. It's height h (in feet) above the ground t seconds after the b
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Question 517795: A ball is thrown straight up with an initial velocity of 80 feet per second from the top of a building 23 feet tall. It's height h (in feet) above the ground t seconds after the ball is thrown is given by h = -16t^2 + 80t + 23. When will the ball be 119 feet above the ground for the first time?
*** ^2 stands for squared or to the second power ***
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
The ball will probably be 119' above the ground twice:
once going up and again coming down.
Just pick the smallest value for and that will
be going up.
Use quadratic formula
and, also,
2 sec from when the ball is thrown, the
height will be 119' above ground for the first time.
check:
OK
Here's a plot:
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