Question

A baseball player throws a baseball with a velocity of 13 m/s North it is caught by a second player seven seconds later how far is the second player from the first player

Answer 1

The second player is 91 meters far from the first player.

Why?

First, let be the +y the North, so, to solve the problem we can use the following formula:

$y=yo+v_o*t+\frac{1}{2}*a*t$

Now, subsituting the given information, we have(assuming that the speed is constant):

\\\\y-yo=13\frac{m}{s}*7s+\frac{1}{2}*0*7s\\\\y-yo=91m\\\\distance=91m[/tex]

Hence, we have that the second player is 91m far from the first player.

Have a nice day!