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 390 m/s2. The ball is launched at an angle of 46° above the ground. Determine the horizontal and vertical components of the launch velocity.
To determine the horizontal and vertical components of the launch velocity, we can use the equations of motion for projectile motion.
Let's break down the given information:
Time of contact (t) = 0.050 s
Acceleration (a) = 390 m/s^2
Launch angle (θ) = 46°
First, we can find the vertical component of the launch velocity (Vy) using the equation:
Vy = a * t
Substituting the values, we get:
Vy = 390 m/s^2 * 0.050 s
Vy = 19.5 m/s
Next, we can find the horizontal component of the launch velocity (Vx) using the equation:
Vx = Vy / tan(θ)
Substituting the values, we get:
Vx = 19.5 m/s / tan(46°)
Vx ≈ 19.5 m/s / 1.03553
Vx ≈ 18.84 m/s
Therefore, the horizontal component of the launch velocity is approximately 18.84 m/s and the vertical component is 19.5 m/s.