Question

A baseball thrown from the outfield is released from shoulder height at an initial velocity of 29.4 m/s at an initial angle of 30.0 with respect to the horizontal. what is the maximum vertical displacement that the ball reaches during its trajectory?

Answer 1

(v^2)=(u^2)+(2*a*s) where v is final speed
u is initial speed
a is acceleration
s is displacement

U=sin30 * 29.4
v=0
a=9.81

Answer 2

22.05 meters

Step-by-step explanation:

In order to get the answer, basically it is a distance we are looking for.

We are going to get familiar with the formula of displacement:

Δx
`=vt`

of the angle of 30°

that is

`29.4(m)/(s^(2)) sin(30)=v_(y) \\ \\29.4(m)/(s^(2))(0.5)=v_(y)\\<strong>14.7(m)/(s)=v_(y)</strong>`average velocity

Now

Δ
`v_(y) =v_(f)-v_(i) \\v_(f)=-14.7 m/s`\\ because the baseball goes the opposite direction when is going down until it gets to the grown. Then
`v_(i)=14.7m/s`

`v_(y) =v_(f)-v_(i)\\v_(y) =-14.7m/s-14.7m/s=-29.4m/s`

and using this formula

`v_(y)=-gt`

we divide both sides by
`-g` and we get
`t`

`-29.4m/s=-9.8m/s^(2)*t \\t=(-29.4m/s)/(-9.8m/s^(2))=3s`

Finally 3 seconds is the time in motion, so we divide by two because we just need the time going up, that happens to be the same time going down.

`(t)/(2)=(3)/(2)=1.5s` time in motion

Δx
`=vt`

`v[/text]= <strong>average velocity=[tex]14.7(m)/(s)`

`t[/text]= <strong>time in motion=[tex]1.5s`

`14.7(m)/(s)*1.5s=22.05m`