An athlete takes part in a long jump competition - Scottish Highers Physics - Question 1 - 2016

To find the horizontal component of the initial velocity, we can use the formula:

$v_{x} = v imes ext{cos}( heta)$

Substituting the values:

To find the time of flight, we can use the following kinematic equation:

$s = ut + \frac{1}{2}at^2$

Here, the vertical motion is influenced by gravity, thus:

• Initial velocity ($u$) is the vertical component we calculated: 3.66 m s⁻¹
• Acceleration ($a$) is due to gravity: -9.8 m s⁻²
• Final displacement is the vertical distance fallen, which we need to solve for.

We can rearrange this equation to find $t$ and considering the 0.76 seconds gives us:

$s = 3.66t + \frac{1}{2}(-9.8)t^2$

At $t = 0.76$ s, if we substitute the values, we can verify that it leads to approximately this time.

The horizontal displacement can be calculated using:

$s_{xh} = v_x t$

Substituting our calculated values:

The graph should show a smaller displacement curve with a decreasing gradient after the initial velocity period, representing the impact of air resistance. Adding this on the displacement vs. time graph will reflect how the displacement diminishes over time compared to the ideal case.