A ball is thrown upward in two dimensions with an initial speed of 17 Zlugs per mega handle(made up) near the surface of the planet earth. What will be the y component of the balls velocity at the peak of its trajectory?

To find the y-component of the ball's velocity at the peak of its trajectory, we can use the principles of projectile motion.

First, let's assume that the positive y-direction is vertically upward. The initial velocity of the ball can be divided into its x-component and y-component. Since the ball is thrown vertically upwards, the x-component of the initial velocity is zero. Therefore, only the y-component needs to be considered.

It is given that the initial speed of the ball is 17 Zlugs per mega handle (which is a made-up unit). Since the ball is thrown vertically upwards, the initial velocity in the y-direction is positive.

Now, we need to determine the time it takes for the ball to reach the peak of its trajectory. In projectile motion, the time it takes for an object to reach its peak can be found using the equation:

t = (Vf - Vi) / g

Where:
t = time taken
Vf = final velocity (at the peak, the vertical component is zero)
Vi = initial velocity (in the y-direction)
g = acceleration due to gravity (approximately 9.8 m/s² on Earth)

Since we know the initial velocity in the y-direction is 17 Zlugs per mega handle (upwards), and the final velocity is 0 at the peak, we can plug the values into the equation:

0 = (0 - 17) / 9.8

Simplifying the equation gives:

0 = -17 / 9.8

Solving for t, we find that the time it takes for the ball to reach the peak of its trajectory is approximately 1.73 seconds.

Now that we have the time, we can find the final y-component of the ball's velocity at the peak of its trajectory. The final velocity in the y-direction can be calculated using the equation:

Vf = Vi - g * t

Where:
Vf = final velocity (at the peak, the vertical component is zero)
Vi = initial velocity (in the y-direction)
g = acceleration due to gravity (approximately 9.8 m/s² on Earth)
t = time taken

Plugging in the known values:

Vf = 17 - (9.8 * 1.73)

Calculating further gives:

Vf ≈ 17 - 16.954 ≈ 0.046

So, the y-component of the ball's velocity at the peak of its trajectory is approximately 0.046 Zlugs per mega handle (upwards).