Respuesta :
Answer:
Vertical distance will be 1.048 m
Explanation:
We have given horizontal speed of the baseball u = 40.2 m/sec
Horizontal distance d = 18.6 m
So time of flight [tex]t=\frac{horizontal\ distance}{speed}=\frac{18.6}{40.2}=0.462sec[/tex]
Now we know that height is given by
[tex]h=ut+\frac{1}{2}gt^2=0\times t+\frac{1}{2}\times 9.8\times 0.462^2=1.048m[/tex]
So vertical distance will be 1.048 m
Vertical distance, when the ball drops as it moves from the pitcher to the catcher is 1.048 meters.
What is the second equation of motion?
The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.
The second equation of the motion for distance can be given as,
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Here, [tex]u[/tex] is the initial velocity, [tex]a[/tex] is the acceleration of the body and [tex]t[/tex] is the time taken by it.
Total time taken by the ball to reach to the catcher is the ratio of horizontal distance traveled by it to the speed of the ball.
As the speed of the ball is 40.2 m/s and the distance of the pitcher and the catcher is 18.6 meters. Thus the time of flight is,
[tex]t=\dfrac{18.6}{40.2}\\t=0.462\rm s[/tex]
Thus, total time taken by the ball to reach to the catcher is 0.462 seconds.
As the initial velocity of the ball is zero. Thus by the second equation of motion,
[tex]h=0+\dfrac{1}{2}(9.81)(0.462)^2\\h=1.048\rm m[/tex]
Thus, the vertical distance covered when the ball drops as it moves from the pitcher to the catcher is 1.048 meters.
Learn more about the equation of motion here;
https://brainly.com/question/13763238
