document.write( "Question 810870: Hi again. I'm sorry to bother you guys again but I have one more question. Here it is: \r
\n" ); document.write( "\n" ); document.write( "A ball is thrown up at 50 m/s at an angle of 37 degrees. How far has the ball traveled (horizontally) when the ball is 25 meters above the ground?\r
\n" ); document.write( "\n" ); document.write( "My AP Physics teacher gave me the answer for this problem, which can be either 39.9 meters OR 205 meters, but she wants me to show how I would get to either of those answers by showing my work. I think the equations I'm using for the problem like v= vi + at to get the time and v = x/t to get the distance are right but the fact that the ball is 25 meters above the ground might be the reason why I can't get either 39.9 m or 205 m. I don't know where 25 meters would fit into any of the equations.
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Algebra.Com's Answer #488521 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
angle of projection = 37 deg\r
\n" ); document.write( "\n" ); document.write( "calculate the time taken to reach a height of 25 m
\n" ); document.write( "The vertical component is v sin \"Theta\"\r
\n" ); document.write( "\n" ); document.write( "y=25=ut+1/2 (-g)t^2 where t is the time in seconds\r
\n" ); document.write( "\n" ); document.write( "25=30t-1/2*9.8t^2\r
\n" ); document.write( "\n" ); document.write( "4.9t^2-30t+25=0\r
\n" ); document.write( "\n" ); document.write( "t= 5.13 & 1\r
\n" ); document.write( "\n" ); document.write( "in the horizontal direction d=rt\r
\n" ); document.write( "\n" ); document.write( "d=v cos\"Theta\"*t
\n" ); document.write( "=v cos 37 *1
\n" ); document.write( "=39.9 meters
\n" ); document.write( "OR
\n" ); document.write( "d=vcos 37 * 5.13
\n" ); document.write( "=204.85 meters
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