In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0771 s, during which time it experiences an acceleration of 162 m/s2. The ball is launched at an angle of 53.3° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.
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To determine the horizontal and vertical components of the launch velocity, we can use the equations of motion for projectile motion.
Given:
Time of contact (t) = 0.0771 s
Acceleration (a) = 162 m/s^2
Launch angle (θ) = 53.3°
(a) Horizontal Component of Launch Velocity:
In projectile motion, there is no horizontal acceleration. Therefore, the horizontal component of velocity remains constant throughout the motion.
The equation for horizontal displacement (d) is given by:
d = Vx * t
Since there is no horizontal acceleration, Vx remains constant. Therefore, we can solve for Vx using the equation:
Vx = d / t
The horizontal displacement is not given in the problem statement, so we cannot determine the exact value of Vx without that information.
(b) Vertical Component of Launch Velocity:
The vertical motion of the ball is affected by the acceleration due to gravity (g), which is approximately 9.8 m/s^2.
The equation for vertical displacement (y) is given by:
y = Vyi * t + (1/2) * a * t^2
At the highest point of the projectile motion, the vertical velocity (Vy) becomes zero.
Using the equation for vertical velocity (Vy), we can find the initial vertical velocity (Vyi):
Vy = Vyi + a * t
0 = Vyi + (-9.8) * t
Solving this equation gives us the value of Vyi.
To find the vertical component of the launch velocity (Viy), we can use the equation:
Viy = Vyi * sin(θ)
Substituting the value of Vyi calculated above, we can find the vertical component of the launch velocity (Viy).
Therefore, without the value of horizontal displacement, we cannot determine the exact horizontal component of the launch velocity. However, we can calculate the vertical component of the launch velocity using the given information.