During what time interval does vy become zero?

So I can't really understand what vy means, I know its a abbreviation but I need help on this problem that has a cannon shooting a cannonball horizontally and it has 3 points; the first being 1s, then 2s, then 3s which is when the ball finally reaches the ground, could you help explain this to me please?

Certainly! In this context, "vy" typically represents the vertical component of velocity. Since the cannonball is shot horizontally, the vertical component of velocity (vy) remains constant throughout its flight. It does not become zero at any point because there is no force acting to change its vertical velocity.

However, it seems like you are trying to determine the time interval during which the cannonball reaches the ground. To do that, we can analyze the vertical motion separately.

Assuming there is no air resistance, we can use the equation for vertical motion:

h = vy * t - (1/2) * g * t^2

where:
- h is the vertical displacement (height)
- t is the time
- vy is the initial vertical velocity (which is typically zero when launched horizontally)
- g is the acceleration due to gravity (which is approximately -9.8 m/s^2)

In this case, since the cannonball reaches the ground at 3 seconds, we can substitute h = 0 and solve for t:

0 = vy * t - (1/2) * g * t^2

Since vy is zero (horizontal launch), the equation simplifies to:

0 = - (1/2) * g * t^2

Simplifying further:

0 = -4.9 * t^2

Now we can solve for t:

t^2 = 0

t = 0 (since the time cannot be negative)

Therefore, the time interval during which the cannonball reaches the ground is from t = 0 seconds to t = 3 seconds.

Of course! I'd be happy to help explain the problem to you.

In this problem, "vy" is most likely representing the vertical component of velocity, specifically the vertical velocity. The letter "v" typically represents velocity, and the subscript "y" indicates that it refers to the vertical direction.

To understand when vy becomes zero, we need to consider the motion of the cannonball. Since the cannonball is shot horizontally, it does not have any initial vertical velocity. However, it is subject to the force of gravity, which causes it to accelerate downwards.

We can use the equation of motion for uniformly accelerated motion in the vertical direction:

y = y₀ + v₀y × t + (1/2)gt²

In this equation, "y" represents the vertical distance traveled by the cannonball at time "t," "y₀" represents the initial vertical position (in this case, the height from where the cannonball was fired), "v₀y" represents the initial vertical velocity, "g" represents the acceleration due to gravity, and "t" represents time.

We know that the cannonball reaches the ground after 3 seconds, so we can substitute this value into the equation. Additionally, since the cannonball is fired horizontally, the initial vertical velocity, v₀y, is zero.

Using these values, the equation becomes:

0 = y₀ + 0 × 3 + (1/2)g(3)²

Simplifying further:

0 = y₀ + (9/2)g

To find the time interval when vy becomes zero, we need to determine when the cannonball reaches its peak height and starts descending. At the peak, the vertical velocity vy will be zero. In other words, vy becomes zero when the cannonball is at the highest point of its trajectory.

From the equation above, we can see that the height y₀ will be negative (since the starting position is above the ground), and the term (9/2)g will be positive due to the acceleration due to gravity.

Therefore, for vy to become zero, the term (9/2)g must be equal in magnitude to y₀ but opposite in sign. This occurs when the cannonball reaches its peak height.

To find the time interval when vy becomes zero, we need more information about the problem, such as the initial height (y₀) from which the cannonball is fired. With that information, we can calculate the time it takes for the cannonball to reach its peak height and determine the interval during which vy becomes zero.

I suggest you repost your COMPLETE problem under the title of physics. I think it will be easier for a physics prof to answer a real rather than a hypothetical question.