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

From our previous calculation, we derived:

$-4.905t^2 + 12.9t - 3 = 0$

Setting this equation to standard form gives us:

$4.91t^2 - 12.9t + 3.00 = 0$

Thus, this confirms that the equation holds.

When solving the equation:

$4.91t^2 - 12.9t + 3.00 = 0,$

we utilize the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 4.91$, $b = -12.9$, and $c = 3.00$. Solving gives two values for $t$, typically one positive and one negative.

1. The positive solution corresponds to the time it takes the ball to pass over the crossbar, which is physically meaningful since time cannot be negative.
2. The negative solution represents a time before the kick, which has no physical significance in this context.

Given the new attempt is from 38 m away with the same initial velocity of 20.0 m s⁻¹, we can use the horizontal motion equation:

$x = ut$

By calculating for $t$ using the horizontal velocity, we find:

$t = \frac{x}{u} = \frac{38}{20 \cos(40^ ext{o})} \approx \frac{38}{15.3}$

While the time taken to reach the crossbar is calculated, we also ensure it doesn’t exceed the earlier calculated time of flight to reach the minimum vertical height. Upon calculation, it shows that the ball will not clear the crossbar.

The gradients in the graphs represent acceleration. Initially, without air resistance, the graph shows a consistent velocity change, indicating constant acceleration.

With air resistance, the gradient becomes less steep over time, signifying a decrease in acceleration as the velocity approaches terminal velocity.

The area under the graph correlates to the vertical distance traveled. In the case without resistance, the greater area indicates a larger distance covered compared to the one with resistance, which shows the effect of environmental factors on projectile motion.