# 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...

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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### 2 Answers

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

=18.3/(2*0.809/9.81)^(1/2)

=45.0605 m/s is the horizontal velocity .