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

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

Answer 2

Answer:

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

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,

$t=\dfrac{d}{v}\\\\t=\dfrac{18.44}{43}\\\\t=0.42\ s$

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 :

$y=\dfrac{1}{2}gt^2\\\\y=\dfrac{1}{2}\times 9.8\times (0.42)^2\\\\y=0.864\ m$

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.