a projectile launched speed of 31.1m/s at angle 71.2 degrees above horizontal fill in table time 0 vx m/s 10.0m/s vy 29.4m/s ax? ay?

Well, let me compute that for you while trying to keep the mood light and humorous!

Let's break it down step by step, shall we?

First, let's find the horizontal velocity (vx) and vertical velocity (vy) components using some trigonometry. We have:

vx = launch speed * cos(angle)
= 31.1 m/s * cos(71.2 degrees)
≈ 10.0 m/s

vy = launch speed * sin(angle)
= 31.1 m/s * sin(71.2 degrees)
≈ 29.4 m/s

So, we've filled in the table with vx = 10.0 m/s and vy = 29.4 m/s.

Now, to determine the horizontal acceleration (ax), we need to know if any external forces are acting on our projectile. Since no external forces were mentioned, we can assume there is no horizontal acceleration (ax = 0 m/s^2). It's just flying through the air, minding its own business!

As for the vertical acceleration (ay), it's affected by gravity. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2, always pulling objects downward. So, we can fill in the table with ay = -9.8 m/s^2. Gravity's got quite a grip on our projectile!

I hope that fills in the blanks for you! If you have any more questions or if you need more silly jokes, feel free to ask!

To fill in the table, we need to calculate the horizontal and vertical components of velocity, as well as the horizontal and vertical components of acceleration.

Given:
Initial speed (Vi) = 31.1 m/s
Launch angle (theta) = 71.2 degrees

Using the given information, we can calculate the horizontal and vertical components of velocity:

Horizontal component of velocity (Vx) = Vi * cos(theta)
= 31.1 * cos(71.2)
≈ 10.0 m/s

Vertical component of velocity (Vy) = Vi * sin(theta)
= 31.1 * sin(71.2)
≈ 29.4 m/s

To calculate the horizontal and vertical components of acceleration, we need to know if there are any external forces acting on the projectile. Without additional information, we assume that there are no horizontal forces acting on the object (ax = 0).

Thus, we can fill in the table as follows:

Time (s) | Vx (m/s) | Vy (m/s) | Ax (m/s^2) | Ay (m/s^2)
------------------------------------------------------
0 | 10.0 | 29.4 | 0 | ?

To fill in the table for the given projectile, we can use the principles of projectile motion. Let's break down the problem step by step:

Step 1: Find the initial horizontal velocity (vx) and initial vertical velocity (vy).
Given:
- Launch speed (magnitude of the initial velocity) = 31.1 m/s
- Launch angle above the horizontal = 71.2 degrees

Using trigonometry, we can calculate the initial horizontal and vertical velocities as follows:

vx = (launch speed) * cos(angle)
vx = 31.1 m/s * cos(71.2 degrees) ≈ 10.0 m/s

vy = (launch speed) * sin(angle)
vy = 31.1 m/s * sin(71.2 degrees) ≈ 29.4 m/s

So, the values for vx and vy are 10.0 m/s and 29.4 m/s, respectively.

Step 2: Find the horizontal acceleration (ax) and vertical acceleration (ay).
In projectile motion, there is no horizontal acceleration (ax). This is because there are no horizontal forces acting on the projectile once it is launched. Therefore, we can write:

ax = 0 m/s^2

The vertical acceleration (ay) is due to the force of gravity and is constant at approximately 9.8 m/s^2. Therefore:

ay = -9.8 m/s^2 (negative sign indicates the acceleration is directed downwards)

So, the values for ax and ay are 0 m/s^2 and -9.8 m/s^2, respectively.

Now, filling in the table:
Time (s) | Vx (m/s) | Vy (m/s) | Ax (m/s^2) | Ay (m/s^2)
------------------------------------------------------
0 | 10.0 | 29.4 | 0 | -9.8

v⒳=v₀•cosα=10.02m/s

v⒴=v₀•sinα=29.44 m/s
a(x)=0
a(y)= -g