You throw a ball upward with an initial speed of 20 m sec About 4 seconds later the returns and you catch it How fast was traveling downward when caught it?

To solve this problem, we can use the equations of motion for a projectile.

The initial velocity of the ball is 20 m/s upward.

The acceleration due to gravity is 9.8 m/s^2 downward.

The time of flight of the ball is 4 seconds.

The velocity of the ball at the top of its trajectory is zero.

The velocity of the ball when it is caught is given by:

```

v = u + at

```

where:

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

Substituting the given values, we get:

```

v = 20 m/s + (9.8 m/s^2)(4 s)

```

```

v = 20 m/s + 39.2 m/s

```

```

v = 59.2 m/s downward

```

Therefore, the ball was traveling downward at a speed of 59.2 m/s when it was caught.