Question

The distance between pitcher and catcher on a baseball field is 18.4 m. If a pitcher releases a ball from a height of 1.5 m and the catcher catches the ball right at ground level, how fast did the pitcher throw the ball?

Answer 1

33.26 m/s

Step-by-step explanation:

The initial velocity of the ball can be determined by;

R = v
`\sqrt{(2H)/(g) }`

Where R is the range (distance between pitcher and catcher), H is the height covered, v is the initial velocity and g is the acceleration due gravity (9.8 m/
`s^(2)`).

So that;

18.4 = v
`\sqrt{(2*1.5)/(9.8) }`

= v
`\sqrt{(3)/(9.8) }`

= v
`√(0.3061)`

= 0.5533v

⇒ v =
`(18.4)/(0.5533)`

= 33.2550

The initial velocity of the ball is 33.26 m/s.This is the velocity with which the pitcher threw the ball.