document.write( "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? \r
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document.write( "I BELIEVE I WOULD USE THE Distance=Rate X Time
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document.write( "Is that right? Is this how I would start it?
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document.write( "112+96t-16t^2=r x t.....i got stuck \n" );
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Algebra.Com's Answer #42418 by Earlsdon(6294) ![]() 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). \n" ); document.write( "You have the appropriate function of distance (really height) as a function of time. \n" ); document.write( " The question here is...when will the ball reach a height (distance) of 112 feet after being thrown upward? \n" ); document.write( "To find out, wou want to set the given function of s(t) equal to 112 ft. and solve for the time, t.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "At t = 0 seconds, the ball is at 112 feet because that the initial height of the throw. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "The ball will be at 112 feet again in 6 seconds. \n" ); document.write( " |