document.write( "Question 413427: An object is thrown upward from a height of 50 ft with an initial velocity of 48 ft per sec. The height in feet after t seconds is given by h=-16t^2+48t+50\r
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document.write( "a. Find the height of the object 3 sec after it is thrown up\r
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document.write( "b. Find the maximum height of the object and the time it takes the object to reach this height \n" );
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Algebra.Com's Answer #290798 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! An object is thrown upward from a height of 50 ft with an initial velocity of 48 ft per sec. The height in feet after t seconds is given by h=-16t^2+48t+50 \n" ); document.write( "a. Find the height of the object 3 sec after it is thrown up \n" ); document.write( "b. Find the maximum height of the object and the time it takes the object to reach this height\r \n" ); document.write( "\n" ); document.write( "..\r \n" ); document.write( "\n" ); document.write( "h=-16t^2+48t+50 \n" ); document.write( "after 3 seconds, \n" ); document.write( "a. h(3)=-16*3^2+48*3+50 \n" ); document.write( " =-144+144+50=50 \n" ); document.write( "What happened is that the object reached its peak then returned to its original height after 3 seconds.\r \n" ); document.write( "\n" ); document.write( "b.You can use the formula, -b/2a to find the time object reaches its maximum height. a=-16, b=48, \n" ); document.write( "Time to reach maximum height =-b/2a=-48/2*-16=-48/-32=3/2 seconds \n" ); document.write( "Maximum height = -16(3/2)^2+(3/2)*48+50=-36+72+50=86 feet\r \n" ); document.write( "\n" ); document.write( "ans: \n" ); document.write( "3 seconds after object is thrown upward, it returns to a height of 50 feet going down. \n" ); document.write( "It takes 1.5 seconds for object to reach its peak after thrown upward. \n" ); document.write( "Maximum height object will reach=86 feet \n" ); document.write( " \n" ); document.write( " |