Asked by Bridgett
Flight of a Ball If a ball is thrown upward at 96 feet per second from the top of a building that is 100 feet high, the height of the ball can be modeled by S(t)=100 + 96t -16t^2 feet, where t is the number of seconds after the ball is thrown. How long after the ball is thrown is the height 228 feet?
228=100+96t-16t^2
228=100+96t-16t^2
Answers
Answered by
GanonTEK
You're almost there. Tidy up the equation to get:
16t^2 - 96t + 128 = 0 which simplifies to
t^2 - 6t + 8 = 0
Now you have a simple quadratic. You'll get two answers because there are two times when it reaches 228ft.
16t^2 - 96t + 128 = 0 which simplifies to
t^2 - 6t + 8 = 0
Now you have a simple quadratic. You'll get two answers because there are two times when it reaches 228ft.
Answered by
ww
S=120+96t-16t^2
Answered by
Ali haider
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