How fast was Ryan's pitch? Answer in units of m/s.
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 m (2.65 ft) by the time it reached home plate, 18.3 m (60 ft) away. The acceleration of gravity is 9.81 m/s^2.
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We can answer this question by looking at the acceleration in the y direction. After the ball leaves Nolan Ryan's hand, the only force acting on it is gravity, in the -y direction.
The equation for distance is d = vt + (1/2)gt^2, where v is the initial velocity of the ball. Since the ball was thrown horizontally, there is no initial downward velocity component to the ball; v = 0.
We know the ball fell .809 m in the time t the ball took to arive at home plate.
.809 = (1/2)(9.81)t^2 --> .165 = t^2
t = 0.41 s
So the ball travelled 18.3 m horizontally in 0.41 s. v = 18.3/0.41 = 44.63 m/s or, 99.83 mph
Note that the official speed for Ryan's pitch is 100.1 mph, thrown in August 1974 at Anaheim Stadium.
The height of the ball when it is thrown horizontally = 0.809 meter.
The horizontal displacement of the ball = 18.3 meter.
Therefore, the time t taken is given by :h= (1/2)gt^2 or t= (2h/g)^(1/2)= 0.4061 secs ---------(1)
So the horizontal velocity =Horizontal displacement/time.
=18.3/time as in (1)
=18.3/[2h/g)^(1/2)] . But h= 0.809m and g=9.81m/s given
=45.0605 m/s is the horizontal velocity .
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