Problema Solution
If a ball is thrown directly upward with a velocity of 40 feet per second, its height (in feet) after t seconds is given by y= 40 t -16 t^2. What is the maximum height attained by the ball?
Answer provided by our tutors
t max=40/16*3
40/32=5/4
puttin value of t in eq. we get
y=40*5/4-16*(\/4)^2
after simlification we get
y=25 which is the maci height.