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 360 m/s2. The ball is launched at an angle of 46° above the ground. Determine the horizontal and vertical components of the launch velocity.

speed = a t starting from rest

speed = 360 * .05

u = speed * cos 46
v = speed * sin 46

To determine the horizontal and vertical components of the launch velocity, we need to analyze the given information and use the equations of motion.

First, let's find the vertical component of the launch velocity.

We know that the ball experiences an acceleration of 360 m/s^2 vertically. The time it remains in contact with the kicker's foot is 0.050 s.

Using the equation of motion,

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken, we can rearrange the equation to solve for the vertical component of the launch velocity:

v = u + at
→ v - at = u

Since the ball starts from rest (u = 0) just before being kicked, the equation simplifies to:

v = at

Plugging in the given values:

v = 360 m/s^2 * 0.050 s
= 18 m/s

Therefore, the vertical component of the launch velocity is 18 m/s.

Now, let's find the horizontal component of the launch velocity.

The horizontal component of the launch velocity remains constant because there is no horizontal acceleration (assuming no air resistance).

To find the horizontal component of the launch velocity, we need to use the equation:

horizontal component of velocity (Vx) = initial velocity (ux) * cos(angle)

The initial velocity (ux) in this case is the total launch velocity, and we can find it using the Pythagorean theorem:

ux = √(Vx^2 + Vy^2)

Substituting the values we already know:

ux = √(0^2 + 18^2)
= √(324)
= 18 m/s

Now, we can find the horizontal component of the launch velocity:

horizontal component of velocity (Vx) = 18 m/s * cos(46°)
= 18 m/s * 0.7193 (rounded to 4 decimal places)

After performing this calculation, the horizontal component of the launch velocity is approximately 12.9465 m/s (rounded to 4 decimal places).

Therefore, the horizontal component of the launch velocity is approximately 12.9465 m/s, and the vertical component of the launch velocity is 18 m/s.