Figure 8 shows the H-shaped posts used in a game of rugby - AQA - A-Level Physics - Question 5 - 2021 - Paper 1

To determine the minimum speed of the ball necessary to just clear the crossbar, we start by breaking down the initial velocity into its horizontal and vertical components.

1. Calculate the vertical component of velocity: The initial vertical velocity can be calculated using:

   $v_{y} = v imes ext{sin}( heta)$

   where $v = 20.0 ext{ m s}^{-1}$ and $heta = 40.0^\circ$. Thus,

   $v_{y} = 20.0 imes ext{sin}(40.0^\circ) \approx 12.9 ext{ m s}^{-1}$

2. Determine the time to reach maximum height: The time taken to reach the maximum height is given by:

   $t_{ ext{up}} = \frac{v_{y}}{g} = \frac{12.9}{9.81} \approx 1.31 ext{ s}$

3. Calculate the vertical displacement to the height of the crossbar: We use the kinematic equation:

   $y = v_{y} t - \frac{1}{2} g t^2$

   Setting $y = 3.00 ext{ m}$ (the height of the crossbar), we can find the minimum required speed at which the ball must be kicked so that it reaches the height of the crossbar just as it passes it:

   $3.00 = 12.9t - \frac{1}{2} imes 9.81 t^2$

   By substituting $t$ back into the equation, it can be shown using quadratic formula that the minimum vertical speed is about 15.0 m s⁻¹ yielding the ball just passes the crossbar.