Area under a basketball shot

Question

The other day, someone asked me how to find the area under a basketball shot. It looked something like this:

How would I go about doing this?

Answer 1

This is also related to physics, and we will ignore minor forces like air friction/drag and other trivial things.

Assumptions: $g$ is the constant of acceleration due to gravity and does not change, aside from the initial force of the ball being thrown and gravity, there are no additional forces, we will consider the ball to be a point-mass object (maybe some others I have forgotten).

The trajectory of the ball is given as:

$$\Delta y=V_{0y}t-gt^2\tag{Galileo's formula}$$

Since gravity does not affect the $x$ direction, and we assume $x_0=0$,

$$x=V_{0x}t$$

$$\implies t=\frac x{V_{0x}}$$

$$\implies\Delta y=V_{0y}\frac x{V_{0x}}-g\frac{x^2}{V_{0x}^2}$$

$$y=V_{0y}\frac x{V_{0x}}-g\frac{x^2}{V_{0x}^2}+y_0$$

The area is simply given as

$$\int_0^dydx=\int_0^dV_{0y}\frac x{V_{0x}}-g\frac{x^2}{V_{0x}^2}+y_0dx$$

$$=V_{0y}\frac d{2V_{0x}}-g\frac{d^3}{3V_{0x}^2}+y_0d$$

where we use metric units and $V_0$ represent the initial velocity in a direction, $g=9.81m/s^2$, $y_0$ is the initial height, and $d$ is the $x$ distance to the hoop.