Suppose I throw a ball straight up at an initial speed of 10 m/s. Suppose I'm 2 meters tall (about 6 feet) and the ball leaves my hand at 2 meters from the ground. How fast is it going at its highest point? How fast is it going on its way back down at 2 meters above the ground. Justify reasoning.

Give me a break. At the highest point its vertical velocity is zero. When it comes back down, it has the same velocity as it started (conservation of energy), but in a downward direction.

To determine the speed of the ball at its highest point and on its way back down, you can use the laws of motion and the principle of conservation of energy.

First, let's calculate the time it takes for the ball to reach its highest point. We can use the equation:

h = ut + (1/2)gt^2

Where:
h = vertical displacement (height)
u = initial velocity (10 m/s upwards)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

Since the ball starts at a height of 2 meters and reaches its highest point, the vertical displacement is 0. The equation becomes:

0 = (10)t - (1/2)(9.8)t^2

Simplifying, we get:

4.9t^2 - 10t = 0

Factoring out t, we have:

t(4.9t - 10) = 0

So, either t = 0 (which is the initial time the ball is thrown) or t = 10/(4.9).

Since the initial time is zero, we can ignore that solution.

Now we can calculate the time it takes for the ball to reach its highest point:

t = 10/(4.9) = 2.04 seconds (approximately)

Now that we know the time it takes for the ball to reach its highest point, we can find the speed at that point. We use the equation:

v = u + gt

Where:
v = final velocity (the speed at the highest point)
u = initial velocity (10 m/s upwards)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (2.04 seconds)

Plugging in the values, we get:

v = 10 + (9.8)(2.04)
v = 10 + 19.992
v = 29.992 ≈ 30 m/s

So, the speed of the ball at its highest point is approximately 30 m/s.

To find the speed of the ball on its way back down when it is 2 meters above the ground, we can use the same formula:

v = u + gt

Plugging in the values:
u = 0 m/s (the ball is briefly at rest at the highest point before falling)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (2.04 seconds)

v = 0 + (9.8)(2.04)
v = 20 ≈ 20 m/s

Therefore, the speed of the ball on its way back down when it is 2 meters above the ground is approximately 20 m/s.

In summary, the ball has a speed of approximately 30 m/s at its highest point and a speed of approximately 20 m/s on its way back down when it is 2 meters above the ground.