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Patrick throws a ball in the air. The height h in feet of the ball after t seconds is h=4+60t-16t^2. After how many seconds is the ball 60 feet above the ground.
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so we should set the height = to 60, and solve for t: \[60 = 4+60t-16t^2 \Rightarrow -56+60t-16t^2 = 0\] you just want to use the quadratic formula to pop out the answer.
sorry but idk how to use that formula sorry
then learn
t=1.75 t=2
\[(-b \pm \sqrt{b^2-4ac})/(2a) \] for when its written as ax^2 + bx+c
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