Initial Statement: A soccer ball is kicked with an initial horizontal velocity of 15 m/s and an initial vertical velocity of 14 m/s.

1) What is the initial speed of the ball?

20.518

2) What is the initial angle θθ of the ball with respect to the ground?

43.025

3) What is the maximum height the ball goes above the ground?

10

4) How far from where it was kicked will the ball land?

42.860

5) What is the speed of the ball 2.2 seconds after it was kicked?

6) How high above the ground is the ball 2.2 seconds after it is kicked?

I've gotten through the first four parts of this problem, but I've gotten stuck on five and six, I think that I need to figure out the time that the ball is at maximum height, which I found was 2.094 seconds, and then use that to find vertical velocity, but everything that I've tried has resulted in a wrong answer.

4. speed of ball at 2 sec. The horizontal speed is same, v(t)=vi-g(r), then compute the speed from those two components.

5. At the top, PE = initial KE in vertical
mgh=1/2 m vi'^2 where vi is initial vertical velocity, solve for h.

To solve questions 5 and 6, you're on the right track by considering the time the ball reaches its maximum height. However, instead of finding the vertical velocity at that time, you can use the known initial vertical velocity.

Let's break it down step by step:

5) To find the speed of the ball 2.2 seconds after it was kicked, you'll need to split the vertical and horizontal motion and treat them separately.

a) Start by finding the horizontal displacement (distance) covered by the ball in 2.2 seconds. Since the horizontal velocity is constant, you can use the formula:

Horizontal displacement = horizontal velocity * time

In this case, the horizontal velocity is 15 m/s, and the time is 2.2 seconds.

b) Next, determine the vertical velocity at 2.2 seconds. The vertical velocity changes due to gravity, so you need to consider the effect of gravity over that time. The vertical velocity at any given time is given by the equation:

Vertical velocity = initial vertical velocity + (acceleration due to gravity * time)

In this case, the initial vertical velocity is 14 m/s, the acceleration due to gravity is -9.8 m/s^2 (negative because it opposes the initial velocity), and the time is 2.2 seconds.

c) Once you have the horizontal and vertical velocities, you can use the Pythagorean theorem to find the magnitude of the total velocity:

Total velocity = √(horizontal velocity^2 + vertical velocity^2)

Plug in the values of the horizontal and vertical velocities you just calculated to find the speed of the ball 2.2 seconds after it was kicked.

6) To find the height of the ball above the ground 2.2 seconds after it is kicked, you can use the formula for vertical displacement (height) under constant acceleration:

Vertical displacement = (initial vertical velocity * time) + (0.5 * acceleration due to gravity * time^2)

Plug in the values of the initial vertical velocity, acceleration due to gravity, and time (2.2 seconds) to find the height above the ground.

By following these steps, you should be able to find the answers to questions 5 and 6 accurately.

To determine the speed of the ball 2.2 seconds after it was kicked, you can use the horizontal velocity and the vertical velocity at that time.

Given:
Initial horizontal velocity (Vx0) = 15 m/s
Initial vertical velocity (Vy0) = 14 m/s
Time (t) = 2.2 s

To find the horizontal speed (Vx) at time t, it remains constant and is equal to the initial horizontal velocity:
Vx = Vx0 = 15 m/s

To find the vertical speed (Vy) at time t, you can calculate the final vertical velocity using the initial vertical velocity and the acceleration due to gravity:
Vy = Vy0 - gt

Where g is the acceleration due to gravity (approximately 9.8 m/s²):

Vy = 14 m/s - (9.8 m/s²)(2.2 s)
Vy = 14 m/s - 21.56 m/s
Vy = -7.56 m/s (negative sign indicates downward direction)

Now, to find the total speed (V) at time t, you can use the horizontal speed (Vx) and vertical speed (Vy):
V = √(Vx² + Vy²)
V = √(15² + (-7.56)²)
V = √(225 + 57.1536)
V = √282.1536
V ≈ 16.796 m/s

Therefore, the speed of the ball 2.2 seconds after it was kicked is approximately 16.796 m/s.

To determine the height above the ground at that time, you can use the initial vertical velocity (Vy0), time (t), and acceleration due to gravity (g):

Height (h) = Vy0 * t + (1/2) * g * t²

h = 14 m/s * 2.2 s + (1/2) * 9.8 m/s² * (2.2 s)²
h = 30.8 m + (1/2) * 9.8 m/s² * 4.84 s²
h = 30.8 m + 47.608 m
h ≈ 78.408 m

Therefore, the ball is approximately 78.408 meters above the ground 2.2 seconds after it is kicked.