Question

A baseball pitcher throws a fastball horizontally at a speed of 43.0 m/s. Ignoring air resistance, how far does the ball drop between the pitcher’s mound and home plate, 60 ft 6 in away?. Express your answer using SI units.

Answer 1

The distance covered by the ball between the pitcher’s mound and home plate is 0.864 meters.

Step-by-step explanation:

Given that,

Horizontal velocity of the football, v = 43 m/s

In this case, we need o find the distance covered by the ball between the pitcher’s mound and home plate, 60 ft 6 in away. It implies,

`d=60\ ft\ 6\ in\\\\d=60.5\ ft\\\\d=8.44\ m`

Let t is the time taken by the football to cover that much distance. So,

`t=(d)/(v)\\\\t=(18.44)/(43)\\\\t=0.42\ s`

Initial velocity of the ball in vertical direction is equal to 0. So, the distance covered by the ball between the pitcher’s mound and home plate is given by :

`y=(1)/(2)gt^2\\\\y=(1)/(2)* 9.8* (0.42)^2\\\\y=0.864\ m`

So, the distance covered by the ball between the pitcher’s mound and home plate is 0.864 meters. Hence, this is the required solution.

Answer 2

Based on your problem where as ask for the distance of the ball drop between the pitchers mound and the home plate and with a given of the speed of ball is 43m/s and the homeplates is 60.6ft away. Based on my step by step procedure and also considering the value of gravity by 9.8m/s^2 i came up with the distance of 144m away