Question

A baseball slugger hits a pitch and watches the ball fly into the bleachers for a home run, landing h = 9.5 m higher than it was struck. When visiting with the fan that caught the ball, he learned the ball was moving with final velocity vf = 30.15 m/s at an angle θf = 29° below horizontal when caught. Assume the ball encountered no air resistance, and use a Cartesian coordinate system with the origin located at the ball's initial position.

Answer 1

The ball's position can get of the initial angle when leave the bat and velocity magnitud.

θ = 35,1

v = 36,5
`(m)/(s)`

Step-by-step explanation:

θf= 29

h= 9,5 m

vf= 30,15
`(m)/(s)`

`V_(ox) =  V_(f) * cos (θ)`

`V_(ox) = 30, 15 * cos ( 29 )`

`V_(ox) = 26,36 (m)/(s)`

Now as you know the vector in 'y' can be related with Tan function

`Tan (θ) = (V_(y) )/(V_(ox) )`

`V_(y) = Tan (29) * 26,36 (m)/(s)`

`V_(y) = 14,617 (m)/(s)`

So, the initial velocity in 'y' can be resolved:

`V_(oy) ^(2)= V_(y) ^(2)  + 2* a * h`

`V_(y) ^(2) = 14,617^(2) +2 * 9,8 (m)/(s^(2)) *9,5 m`

`V_(oy) = √(399,8569828)`

`V_(oy) =19,99642525 (m)/(s)`

Finally the velocity is going to be:

`V= \sqrt{V_(ox) ^(2) +V_(oy) ^(2)  }`

`V= \sqrt{26,36 ^(2) +19,99 ^(2)  }`

`V= 33,08 (m)/(s)`

θ
`= Tan^(-1) ((19,99)/(26,36))`

θ
`=37,188`