A ball thrown vertically...

Question

A ball is thrown vertically upward from the top of a building 80 feet tall with an initial velocity of 64 ft per second. The distance s in feet of the ball from the ground after T seconds is s equals 80 + 64t- 16t^2. After how many seconds does the ball strike the ground ? after how many seconds will the ball past the top of the building on its way down?

Answer 1

To find when it strikes the ground, set

 

s(t) = 0

 

and solve for t using the quadratic formula.  You will get a negative and a positive solution.  Discard the negative solution, since time starts at t = 0.

 

To find when it passes the top of the building on the way down, set

s(t) = 80,

 

simplify the expression, and solve for t using the quadratic formula. You will get zero and a positive solution. The positive solution is the time it passes 80 ft on the way down.

 

You now have everything you need to solve the problem.  Finish from here.