Answer:
The velocity of the baseball, rounded to the nearest hundredth, is 0.82 m/s.
Explanation:
We will use equation of motion:
[tex]h=ut+\frac{1}{2}gt^2[/tex]
h is the initial height
u = 0 = Initial vertical velocity
[tex]g = 9.8 m/s^2[/tex] = Acceleration of gravity
t = Time
We are given that A baseball is launched horizontally from a height of 1.8 m.
So, Substitute h = 1.8 m
[tex]1.8=(0)t+\frac{1}{2}(9.8) t^2\\3.6=(9.8) t^2\\\frac{3.6}{9.8}=t^2\\\sqrt{\frac{3.6}{9.8}}=t\\0.61=t[/tex]
We are given that The baseball travels 0.5 m before hitting the ground.
Now Distance = 0.5 m
Time = 0.61 sec
To Find Speed
[tex]Speed = \frac{Distance}{Time}\\Speed = \frac{0.5}{0.61}\\Speed=0.819[/tex]
So, Speed = 0.82 m/s
Hence The velocity of the baseball, rounded to the nearest hundredth, is 0.82 m/s.
