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4. A rugby ball is kicked during and it leaves the ground at an angle of 28 degrees with a velocity of 60.0 m/s. Assume the ball leaves the foot at ground level and ignore air resistance.

4.1. Determine the x and y components of the initial velocity.
4.2. Calculate the maximum height reached by the ball.
4.3. Calculate the time taken by the ball to reach maximum height.
4.4. Calculate the time of travel before the football hits the ground.
4.5. Calculate how far away the ball hits the ground.

Answer :

This question involves concepts related to projectile motion in physics, specifically involving a rugby ball kicked at an angle. Let's break down each part:

4.1. Determine the x and y components of the initial velocity.

To find the components of the initial velocity, we use trigonometric functions:

  • The x-component (horizontal) of the velocity is given by:
    [tex]v_x = v \cdot \cos(\theta)[/tex]
    where [tex]v = 60.0 \, \text{m/s}[/tex] and [tex]\theta = 28^\circ[/tex].

    [tex]v_x = 60.0 \times \cos(28^\circ) \approx 60.0 \times 0.8829 \approx 52.97 \, \text{m/s}[/tex]

  • The y-component (vertical) of the velocity is given by:
    [tex]v_y = v \cdot \sin(\theta)[/tex]

    [tex]v_y = 60.0 \times \sin(28^\circ) \approx 60.0 \times 0.4695 \approx 28.17 \, \text{m/s}[/tex]

4.2. Calculate the maximum height reached by the ball.

The formula for the maximum height [tex]h_{max}[/tex] in projectile motion, assuming upward velocity only, is:

[tex]h_{max} = \frac{{v_y^2}}{{2g}}[/tex]

where [tex]g = 9.81 \, \text{m/s}^2[/tex] is the acceleration due to gravity.

[tex]h_{max} = \frac{{(28.17)^2}}{{2 \times 9.81}} \approx \frac{{793.0689}}{{19.62}} \approx 40.42 \, \text{m}[/tex]

4.3. Calculate the time taken by the ball to reach maximum height.

The time to reach the maximum height [tex]t_{max}[/tex] can be calculated by:

[tex]t_{max} = \frac{{v_y}}{{g}}[/tex]

[tex]t_{max} = \frac{{28.17}}{{9.81}} \approx 2.87 \, \text{seconds}[/tex]

4.4. Calculate the time of travel before the ball hits the ground.

The total time of flight [tex]T[/tex] for a projectile launched from and landing on the same horizontal level is:

[tex]T = 2 \times t_{max}[/tex]

[tex]T = 2 \times 2.87 \approx 5.74 \, \text{seconds}[/tex]

4.5. Calculate how far away the ball hits the ground.

The horizontal distance [tex]R[/tex] (also called the range) is given by:

[tex]R = v_x \times T[/tex]

[tex]R = 52.97 \times 5.74 \approx 304.044 \, \text{m}[/tex]

These calculations show how the components of the initial velocity are used to determine other characteristics of projectile motion, like maximum height, time, and range.

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Rewritten by : Barada