A boy tosses a football upward. If the football rises a vertical distance of 1.3 m and the boy catches it at the same point he released it, what is the velocity of the ball just before he catches it?

A boy tosses a football upward. If the football rises a vertical distance of 3.7 m and the boy catches it at the same point he released it, what is the velocity of the ball just before he catches it?

To find the velocity of the ball just before the boy catches it, we can use the kinematic equations of motion.

First, let's analyze the motion of the ball. Since the ball is tossed upward and then caught at the same point, we can assume that its initial and final vertical positions are the same. Therefore, we can use the following equation:

v^2 = u^2 + 2as

Where v is the final velocity of the ball, u is the initial velocity of the ball, a is the acceleration, and s is the displacement.

In this case, the displacement (s) is given as 1.3 m, and the ball is moving in the opposite direction to gravity. We know that the acceleration due to gravity is approximately -9.8 m/s². Therefore, we can substitute these values into the equation:

v^2 = 0 + 2(-9.8)(1.3)

Simplifying, we have:

v^2 = -25.48

Since we are interested in the velocity just before he catches the ball, we consider the magnitude of the velocity. Taking the square root of both sides of the equation, we find:

v = √25.48

v ≈ 5.05 m/s

So, the velocity of the ball just before the boy catches it is approximately 5.05 m/s.