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 380 m/s2. The ball is launched at an angle of 51° above the ground. Determine the horizontal and vertical components of the launch velocity.
Vx= ____ m/s
Vy= ____ m/s
To solve this problem, we can use the equations of motion to calculate the horizontal and vertical components of the launch velocity.
First, let's analyze the vertical motion. We can use the equation of motion to relate the vertical displacement, initial vertical velocity, time, and acceleration:
Δy = Vyi * t + (1/2) * a * t^2
Since the ball is launched at an angle of 51° above the ground, we know that the vertical component of the initial velocity, Vyi, is given by:
Vyi = V * sin(θ)
Where V is the magnitude of the launch velocity and θ is the angle of the launch (51° in this case).
Now, we can use the given information to solve for the vertical component:
Δy = 0 (since we're assuming the ball lands at the same height it was launched from)
Vyi = V * sin(θ) = Vy
t = 0.050 s
a = 380 m/s^2
0 = Vy * 0.050 s + (1/2) * 380 m/s^2 * (0.050 s)^2
Simplifying the equation gives us:
0 = Vy * 0.050 s + (1/2) * 380 m/s^2 * 0.0025 s^2
0 = Vy * 0.050 s + 0.2375 m
-0.2375 m = Vy * 0.050 s
Therefore:
Vy = -0.2375 m / 0.050 s
Vy = -4.75 m/s
The negative sign indicates that the vertical component is directed downwards.
Now, let's analyze the horizontal motion. The horizontal component of the initial velocity, Vxi, is given by:
Vxi = V * cos(θ)
Next, we need to determine the time taken by the ball to remain in contact with the kicker's foot. This time, 0.050 s, is the same for both horizontal and vertical motion since the ball leaves the foot simultaneously in both directions.
Now, the horizontal displacement, Δx, can be calculated using the equation of motion:
Δx = Vxi * t
Since there is no external force acting horizontally, the acceleration, ax, is 0 m/s^2. This means that the horizontal velocity remains constant throughout the motion.
Substituting the given information into the equation gives us:
Δx = Vxi * 0.050 s
Simplifying gives us:
Δx = Vxi * 0.050 s
Δx = V * cos(θ) * 0.050 s
Therefore:
Vx = Δx / 0.050 s
Vx = V * cos(θ) * 0.050 s / 0.050 s
Vx = V * cos(θ)
Now, we can substitute the given angle:
Vx = V * cos(51°)
Since we do not have the value of V, we cannot determine the exact values of Vx and Vy. However, we can calculate their relative magnitudes and ratios.
To summarize:
- Vx = V * cos(51°)
- Vy = -4.75 m/s