Question

6. In baseball, a fastball takes about 1/2 second to reach the plate. Assuming that such a pitch is thrown horizontally, compare the distance the ball falls in the first 1/4 second with the distance it falls in the second 1/4 second.

Answer 1

Answer:

The ball falls 0.3 m in the first 1/4 seconds, and it falls 0.92 m in the second 1/4 second

Explanation:

Horizontal Launch

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.

To calculate the vertical distance (y) traveled by the object in terms of the time (t) we use:

$\displaystyle y=\frac{g.t^2}{2}$

The fastball is thrown horizontally. The distance the ball falls in t=1/4 seconds = 0.25 seconds, is:

$\displaystyle y_1=\frac{9.8*0.25^2}{2}$

$\displaystyle y_1=0.30625\ m$

The distance the ball falls in t=1/2 seconds = 0.5 seconds is:

$\displaystyle y_2=\frac{9.8*0.5^2}{2}$

$\displaystyle y_2=1.225\ m$

The distance it falls in the second 1/4 second is:

$y_2-y_1=1.225\ m-0.30625\ m$

$y_2-y_1=0.91875\ m$

The ball falls 0.3 m in the first 1/4 seconds, and it falls 0.92 m in the second 1/4 second