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!
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
To find the speed of the pitch, we can use the kinematic equation for displacement in the y-direction:
Δy = v₀y * t + (1/2) * a * t²
In this equation:
- Δy is the change in y displacement (-0.809m in this case)
- v₀y is the initial velocity in the y-direction (which we'll solve for)
- a is the acceleration in the y-direction (which is -9.81m/s², representing the acceleration due to gravity)
- t is the time taken for the ball to reach home plate.
Since the pitch is thrown horizontally, the initial velocity in the y-direction is 0. Therefore, the equation becomes:
Δy = (1/2) * a * t²
Rearranging the equation, we can solve for t:
t² = (2 * Δy) / a
t = √((2 * Δy) / a)
Substituting the given values, we get:
t = √((2 * -0.809) / -9.81)
Calculating this expression, we find t ≈ 0.128 seconds.
Now, to find the speed of the pitch, we can utilize the kinematic equation for displacement in the x-direction:
Δx = v₀x * t
Since the pitch is thrown horizontally, the initial velocity in the x-direction (v₀x) is the speed we're looking for. Rearranging the equation, we get:
v₀x = Δx / t
Substituting the given values, we have:
v₀x = 18.3 / 0.128
Calculating this expression, we find v₀x ≈ 142.97 m/s.
Therefore, the pitch was thrown at a speed of approximately 142.97 meters per second.