# SOLUTION: A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground af

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Travel -> SOLUTION: A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground af      Log On

 Ad: Over 600 Algebra Word Problems at edhelper.com Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Word Problems: Travel and Distance Solvers Lessons Answers archive Quiz In Depth

 Question 61621: A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=112+96t-16t^2. After how many seconds will the ball pass the top of the building on its way down? I BELIEVE I WOULD USE THE Distance=Rate X Time Is that right? Is this how I would start it? 112+96t-16t^2=r x t.....i got stuckAnswer by Earlsdon(6287)   (Show Source): You can put this solution on YOUR website!The use of the distance formula, d = rt, is not appropriate for this kind of problem. The reason is the distance formula applies in cases of constant motion (speed) whereas this problem involves acceleration (due to gravity). You have the appropriate function of distance (really height) as a function of time. The question here is...when will the ball reach a height (distance) of 112 feet after being thrown upward? To find out, wou want to set the given function of s(t) equal to 112 ft. and solve for the time, t. Subtract 112 from both sides. Factor out a t. Apply the zero product principle. and At t = 0 seconds, the ball is at 112 feet because that the initial height of the throw. Add 16t to both sides. Divide both sides by 16. The ball will be at 112 feet again in 6 seconds.