set s(t) to 96 and solve.
96=96+80t-16t^2
solve for t
A ball is thrown upward from the top of a building which is 96ft tall w/ Initial velocity of 80 ft per sec.
Distance s(ft) of ball from the ground after t sec is s=96+80t-16t^(2)
1) how many sec. will the ball pass the top of the building when it is coming down?
How would I solve this?
5 answers
why would I set 96 for s(t) because that's the height of the building ?
but it's when the ball passes the top of the building
Also I tried solving it I got to
-16t(t-5)= 0 what should I do next
but it's when the ball passes the top of the building
Also I tried solving it I got to
-16t(t-5)= 0 what should I do next
I finished solving it. thank you
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A ball is thrown vertically upward with velocity 100ft/sec and height of building is 40ft what is maximum height reached by ball and what is velocity when ball hits the ground