Question

A baseball pitcher throws a ball horizontally at a speed of 40.2 m/s. A catcher is 18.6 m away from the pitcher. Find the magnitude, in meters, of the vertical distance that the ball drops as it moves from the pitcher to the catcher. Ignore air resistance.

Answer 1

Answer:

Vertical distance will be 1.048 m

Explanation:

We have given horizontal speed of the baseball u = 40.2 m/sec

Horizontal distance d = 18.6 m

So time of flight $t=\frac{horizontal\ distance}{speed}=\frac{18.6}{40.2}=0.462sec$

Now we know that height is given by

$h=ut+\frac{1}{2}gt^2=0\times t+\frac{1}{2}\times 9.8\times 0.462^2=1.048m$

So vertical distance will be 1.048 m