Abaseball 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

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:

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!

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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