Question

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

Answer 1

Answer:

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

Explanation:

We will use equation of motion:

$h=ut+\frac{1}{2}gt^2$

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

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

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

$Speed = \frac{Distance}{Time}\\Speed = \frac{0.5}{0.61}\\Speed=0.819$

So, Speed = 0.82 m/s

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