document.write( "Question 1178129: a person throws a ball upward into the air with an initial velocity of 15.0 m/s.
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document.write( "a) calculate how high it goes
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document.write( "b)calculate how long the ball is in the air before it comes back to his hand.
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document.write( "c) how much time it takes for the ball to reach the maximum height?
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document.write( "d) find the velocity of the ball when it returns to the thrower's hand \n" );
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Algebra.Com's Answer #807296 by mananth(16946)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Here acceleration (a) is g = 9.8m/s^2 (this will be negative since it is going in the opposite direction of the initial velocity),\r \n" ); document.write( "\n" ); document.write( "initial velocity(u) = 15m/s \n" ); document.write( "final velocity (v) = 0m/s if the ball stops \n" ); document.write( "v^2 = u^2 + 2a*h, where h is the max height of the ball, \n" ); document.write( " we can find h by solving: h = (v^2 - u^2)/2a, \n" ); document.write( " we plug in h = (0^2 - 15^2)/2(-9.8) = 11.48 m.\r \n" ); document.write( "\n" ); document.write( "b)calculate how long the ball is in the air before it comes back to his hand.\r \n" ); document.write( "\n" ); document.write( " time to reach its peak and back down to our hand. \n" ); document.write( " Using this, the time it takes to get to the top * 2 = total time of flight . Using d = (v + u)*t / 2, ) = 2(11.48)/(15) = 1.53 s to reach top,\r \n" ); document.write( "\n" ); document.write( "we get a total of2*1.53 = 3.06 s to go up and back down to our hand. \n" ); document.write( " \n" ); document.write( " |