In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.050 s, during which time it experiences an acceleration of 340 m/s2. The ball is launched at an angle of 51° above the ground. Determine the horizontal and vertical components of the launch velocity.

Since average a= delta v/ delta t

340(.05)= delta v
v=17 m/s
Furthermore this launch velocity has the angle of 51 degrees. You then use your handy dandy trigonometric skills to find the horizontal and vertical components of the launch velocity.

17cos(51)= horizontal
find the vertical?

13.21

To determine the horizontal and vertical components of the launch velocity, we first need to find the initial velocity of the ball.

Given:
Acceleration (a) = 340 m/s^2
Time (t) = 0.050 s
Launch Angle (θ) = 51°

Step 1: Find the initial velocity in the vertical direction (Vyi):

Using the equation:
Vyi = a * t

Vyi = 340 m/s^2 * 0.050 s = 17 m/s

Step 2: Find the initial velocity in the horizontal direction (Vxi):

Using the equation:
Vxi = V * cos(θ), where V is the initial velocity

We need to find V first, using the equation:
V = a * t

V = 340 m/s^2 * 0.050 s = 17 m/s

Now, substitute this value into the equation for Vxi:
Vxi = 17 m/s * cos(51°) ≈ 10.91 m/s

So, the horizontal component of the launch velocity is approximately 10.91 m/s, and the vertical component is approximately 17 m/s.

To determine the horizontal and vertical components of the launch velocity, we can use the equations of motion.

First, let's define the given information:
Time of contact with the kicker's foot (t) = 0.050 s
Acceleration (a) = 340 m/s^2
Launch angle (θ) = 51°

Now, we can find the vertical component (Vy) and horizontal component (Vx) of the launch velocity.

Step 1: Find the vertical component (Vy):
We know that the only force acting on the ball during contact with the kicker's foot is gravity. Therefore, the vertical acceleration is equal to the acceleration due to gravity (9.8 m/s^2), downward.

Using the equation of motion for vertical displacement, we have:
Sy = Voy * t + (1/2) * a * t^2
Since the ball starts from rest vertically, Voy = 0.

Therefore, Sy = (1/2) * a * t^2.

Substituting the given values:
Sy = (1/2) * (9.8 m/s^2) * (0.050 s)^2.

Simplifying this equation gives us the vertical displacement.

Step 2: Find the horizontal component (Vx):
The horizontal acceleration is zero because no horizontal force is acting on the ball. Therefore, the horizontal component of the velocity, Vx, remains constant throughout the motion.

Using the equation of motion for horizontal displacement, we have:
Sx = Vx * t.

But since the horizontal acceleration is zero, Sx = Vx * t simplifies to: Sx = Vx * (0.050 s).

Now, we can rearrange the equation to solve for Vx:
Vx = Sx / (0.050 s).

Step 3: Calculate the values of Sy and Sx:
To find Sy and Sx, we need to determine the total displacement of the ball in the vertical and horizontal directions.

Using the equations of projectile motion, we have:
Sy = Voy * t + (1/2) * a * t^2, and
Sx = Vx * t.

Since the ball is launched at an angle, we need to find the vertical and horizontal velocities (Voy and Vx) in terms of the launch velocity (V) and launch angle (θ).

Voy = V * sin(θ),
Vx = V * cos(θ).

Step 4: Calculate Vy and Vx:
Using the given launch angle (θ), we can find the values of Voy and Vx:

Voy = V * sin(θ) = V * sin(51°),
Vx = V * cos(θ) = V * cos(51°).

Substituting these equations into steps 2 and 3, we can solve for Sy and Sx.

Step 5: Solve for Sy and Sx:
Sy = (1/2) * (9.8 m/s^2) * (0.050 s)^2,
Sx = (V * cos(51°)) * (0.050 s).

Step 6: Solve for Vx:
Vx = Sx / (0.050 s).

By substituting the given values into these equations, you can find the vertical and horizontal components of the launch velocity.