The ball is thrown horizontally with some speed so that it will fall vertically downwards by distance y = 0.809 m
now using kinematics we can find the time to fall such distance
[tex]y = \frac{1}{2}gt^2[/tex]
[tex]0.809 = \frac{1}{2}\times 9.8 \times t^2[/tex]
[tex]t = 0.406 s[/tex]
now in the same time it will reach to the distance of 18.3 m in horizontal direction
So here we will have
[tex]v = \frac{d}{t}[/tex]
[tex]v = \frac{18.3}{0.406} = 45.0 m/s[/tex]
So it is travelling at 45.0 m/s
The ball will reach the home plate at the speed of 45 m/s.
Given data:
Distance covered by ball is, d = 0.809 m.
Distance between the runner and home plate is, d' = 18.3 m.
Applying the second kinematic equation of motion to find the time taken by ball to reach the ground as,
[tex]d=ut+\dfrac{1}{2}gt^{2} \\0.809=0 \times t+\dfrac{1}{2} \times 9.8t^{2} \\t=\sqrt{\dfrac{2 \times 0.809}{9.8} }\\t = 0.406 \;\rm s[/tex]
Now, the speed of ball to reach the home plate is given as,
[tex]v = \dfrac{d'}{t} \\v = \dfrac{18.3}{0.406}\\v=45.0 \;\rm m/s[/tex]
Thus, the ball will reach the home plate at the speed of 45 m/s.
Learn more about kinematics equation of motion here:
https://brainly.com/question/13202578?referrer=searchResults
