A football is thrown horizontally with an initial velocity of (16.5 m/s). Ignoring air resistance, the average acceleration of the football over any period of time is (-9.81 m/s2). Find the magnitude and direction of the velocity at this time and direction (as an angle) below the horizon .

I am not going to do this for you. The horizontal velocity is constant, the vertical velocity increases due to gravity. The final velocity is a resultant of the horizontal and vertical.

I will be happy to critique your work.

Oh, I see you're trying to tackle a physics problem! Well, let me lend you a hand, or should I say a joke.

Why did the football bring a ladder to the game? Because it wanted to reach new heights!

Now let's get to the problem. Since the acceleration due to gravity is acting vertically, it won't affect the horizontal velocity. Therefore, the magnitude of the horizontal velocity remains constant at 16.5 m/s.

Now, when it comes to the vertical velocity, it's constantly changing due to gravity. So, we need to determine its magnitude and direction. Since the acceleration due to gravity is -9.81 m/s^2, this means the vertical velocity increases downwards.

To find the magnitude of the resultant velocity, we can use the Pythagorean theorem. The horizontal and vertical velocities form a right triangle, with the resultant velocity as the hypotenuse.

Let me quickly calculate that for you:
magnitude = sqrt((horizontal velocity)^2 + (vertical velocity)^2)
magnitude = sqrt((16.5 m/s)^2 + ((-9.81 m/s^2) * t)^2)

As for the direction, you mentioned it should be below the horizon. Well, since the vertical velocity is directed downwards, we can say the direction is negative in the vertical direction.

Now, since I've given you the tools, it's time for you to do the math! Good luck, and remember to keep your eyes on the ball (or the numbers)!

And if you need any more jokes or assistance, feel free to ask!

To find the magnitude and direction of the velocity at this time below the horizon, we can break down the problem into its horizontal and vertical components.

Given:
Initial horizontal velocity (u) = 16.5 m/s
Average acceleration (a) = -9.81 m/s^2 (directed downward)

1. Horizontal motion:
The horizontal velocity remains constant throughout the motion. Therefore, the magnitude of the horizontal velocity at any point in time remains the same as the initial velocity.

Magnitude of horizontal velocity (v_x) = u = 16.5 m/s

2. Vertical motion:
Since the only force acting on the football in the vertical direction is gravity, the vertical velocity increases uniformly with time.

Using the kinematic equation for vertical motion:
v_y = u_y + at

Where:
v_y = final vertical velocity
u_y = initial vertical velocity (which is 0, since the ball is thrown horizontally)
a = average vertical acceleration (gravity)

Substituting the given values:
v_y = 0 + (-9.81 m/s^2) = -9.81 m/s

The magnitude of the final velocity (v) can be found using the Pythagorean theorem:

v = √(v_x^2 + v_y^2)
v = √((16.5 m/s)^2 + (-9.81 m/s)^2)

Calculating the magnitude of the final velocity (v):
v ≈ 18.60 m/s

To find the direction of the velocity below the horizon (as an angle), we can use the tangent inverse function:

θ = tan^(-1)(v_y / v_x)
θ = tan^(-1)(-9.81 m/s / 16.5 m/s)

Calculating the direction of the velocity (θ):
θ ≈ -29.28 degrees

Therefore, at this time below the horizon, the magnitude of the velocity is approximately 18.60 m/s and it is directed at an angle of approximately -29.28 degrees below the horizon.

To find the magnitude and direction of the velocity at a specific time, you can use the principles of projectile motion.

Step 1: Break down the initial velocity into horizontal and vertical components. Since the football is thrown horizontally, the horizontal component will be equal to the initial velocity (v₀) given as 16.5 m/s. The vertical component is 0 m/s initially because there is no initial vertical velocity.

Step 2: Determine the time at which you want to find the velocity. Let's denote this time as t.

Step 3: Calculate the vertical displacement using the equation:
Δy = v₀yt + (1/2)at², where v₀y is the vertical component of the initial velocity, a is the acceleration (-9.81 m/s²), and t is the time. Since the initial vertical component is 0 m/s, the equation simplifies to Δy = (1/2)at².

Step 4: Calculate the vertical velocity at time t using the equation:
v_y = v₀y + at, where v₀y is the vertical component of the initial velocity and a is the acceleration. Again, since the initial vertical component is 0 m/s, this equation simplifies to v_y = at.

Step 5: Calculate the magnitude of the resultant velocity using the Pythagorean theorem:
v = √(v_x² + v_y²), where v_x is the horizontal component of the initial velocity (16.5 m/s) and v_y is the vertical velocity calculated in step 4.

Step 6: Calculate the direction of the resultant velocity as an angle below the horizon. You can use the inverse tangent function:
θ = tan⁻¹(v_y / v_x), where v_x is the horizontal component of the initial velocity (16.5 m/s) and v_y is the vertical velocity calculated in step 4.

By following these steps, you will be able to find the magnitude and direction of the velocity at a specific time below the horizon.