Question

A baseball is launched horizontally from a height of 1.8 m. The baseball travels 0.5 m before hitting the ground.

The velocity of the baseball, rounded to the nearest hundredth, is ____ m/s.

Answer 1

0.82

Step-by-step explanation:

Answer 2

The velocity of the baseball, rounded to the nearest hundredth, is 0.82 m/s.

Step-by-step explanation:

We will use equation of motion:

`h=ut+(1)/(2)gt^2`

h is the initial height

u = 0 = Initial vertical velocity

`g = 9.8 m/s^2` = 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

`1.8=(0)t+(1)/(2)(9.8) t^2\\3.6=(9.8) t^2\\(3.6)/(9.8)=t^2\\\sqrt{(3.6)/(9.8)}=t\\0.61=t`

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

`Speed = (Distance)/(Time)\\Speed = (0.5)/(0.61)\\Speed=0.819`

So, Speed = 0.82 m/s

Hence The velocity of the baseball, rounded to the nearest hundredth, is 0.82 m/s.