Question

A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizontally from an elevation 5.0 mm above the ground. The ball lands 20 mm away. What is his pitching speed?

Answer 1

`v_(ox)= 19.6\ m/s`

Step-by-step explanation:

Data provided in the question:

Height above the ground, H= 5.0m

Range of the ball, R= 20 m

Initial horizontal velocity =
`v_(ox)`

Initial vertical velocity=
`v_(oy)` (Since ball was thrown horizontally only)

Acceleration acting horizontally,
`a_x` = 0 m/s² [ Since no acceleration acts horizontally) ]

Vertical Acceleration,
`a_y` = 9.8 m/s² (Since only gravity acts on it)

Let 't' be the time taken to reach ground

Therefore, using equations of motion, we have

`H= v_(oy)t+(1)/(2)a_yt^2`

`5= (0)t+(1)/(2)(9.8)t^2`

`t= (10)/(9.8)=1.02 s`

Then using Equations of motion for horizontal motion,

`R= v_(ox)t+(1)/(2)a_xt^2`

`20= v_(ox)(1.02)+(1)/(2)(0)(1.02)^2`

`v_(ox)= 19.6\ m/s`