Question

The fastest recorded pitch in Major League Baseball was thrown by Nolan Ryan in 1974. If this pitch were thrown horizontally, the ball would fall 0.809 meters by the time it reached home plate, 18.3 meters away. How fast was the pitch?

So for i know that change in y displacement is -.809 and acceleration of y is -9.81. Also change in x displacement is 18.3m and initial velocity i think is 0.

I can't seem to put it all together to find the answer, help please!

Answer 1

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

Answer 2

Would you mind explaining how you got the answer in words please? - would really appreciate it.

Answer 3

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