Posted by Michael on Saturday, October 13, 2012 at 10:05am.
x=18.3 m; y=0.809 m. v(x) = ?
x=v(x) •t ,
x²=v(x)² •t² ,
y=g•t²/2 ,
x²/y= 2•v(x)² •t²/g•t²=2•v(x)² /g
v(x)=sqrt{ x²•g/2•y} =sqrt{18.3²•9.81/2•0.809}=12.95 m/s
Would you mind explaining how you got the answer in words please? - would really appreciate it.
The ball is horizontal projectile thrown with velocity v(x). It takes part in two motions simultaneously: along the horizontal and vertical axis. The horizontal motion of the projectile is uniform motion x=v(x)•t. This is because the only force acting on the projectile is force of gravity: the vertical motion is accelerated motion with initial velocity v(oy) = 0 and acceleration ‘g’ => y=g•t²/2.
For solution, I squared the first equation x²=v(x)² •t² , and then the result I divided by the second equation x²/y= 2•v(x)² •t²/g•t²=2•v(x)² /g. Solving for ‘v(x), I obtained
v(x)=sqrt{ x²•g/2•y} =sqrt{18.3²•9.81/2•0.809}=12.95 m/s
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