In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0854 s, during which time it experiences an acceleration of 217 m/s2. The ball is launched at an angle of 58.3° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.
Vo=a*t = 217 * 0.0854 = 18.53 m/s[58.3o]
a. Xo = 18.53*Cos58.3.
b. Yo = 18.53*sin58.3
To determine the horizontal and vertical components of the launch velocity, we can use the given information on the duration of contact, acceleration, and launch angle.
(a) The horizontal component of the launch velocity remains constant throughout the motion, as there is no external force acting horizontally. Therefore, the horizontal component of the velocity is given by the formula:
Vx = (distance) / (time)
Since there is no information given about the distance, we cannot directly calculate the horizontal component of the velocity. However, we can assume that it remains constant over the entire duration of contact. Therefore, if we have the initial horizontal velocity (Vx0), it will be equal to the final horizontal velocity (Vx).
(b) The vertical component of the launch velocity can be calculated using the equations of motion. We can use the following kinematic equation:
Vf = Vi + at
Where:
Vf is the final vertical velocity
Vi is the initial vertical velocity
a is the acceleration due to gravity (-9.8 m/s^2)
t is the time of contact (0.0854 s)
At the highest point of the ball's trajectory, the vertical velocity becomes zero. Using this information, we can calculate the initial vertical velocity (Vi). Rearranging the equation, we get:
Vi = Vf - at
Since Vf = 0 at the highest point, we can simplify the equation:
Vi = -at
Now substituting the given values, we have:
Vi = -9.8 m/s^2 * 0.0854 s = -0.83692 m/s
Therefore, the initial vertical velocity is -0.83692 m/s. Notice the negative sign indicates that the velocity is directed opposite to the gravitational acceleration.
To find the initial launch velocity, we can use the given launch angle and the components of the velocity:
Vx0 = V * cosθ
Vy0 = V * sinθ
Where:
Vx0 is the initial horizontal velocity
Vy0 is the initial vertical velocity
V is the initial launch velocity
θ is the launch angle (58.3°)
We can rearrange these equations to solve for V:
V = √(Vx0^2 + Vy0^2)
Substituting the values we have:
V = √(Vx0^2 + (-0.83692 m/s)^2)
Since we do not have a specific value for the horizontal velocity, we cannot calculate the exact launch velocity. However, we can calculate the magnitude of the initial launch velocity by assuming Vx0 = Vx (horizontal component):
V = √(Vx^2 + (-0.83692 m/s)^2)
This equation will give you the magnitude of the initial launch velocity.