Launching from a Cut Parabola

Relevant wiki: 2D Kinematics Problem-solving

$Let :$

• m = the mass of the bead ;
• g = the gravitational acceleration of the Earth ;

Since the only force acting on the bead is gravity we can write down that the difference between the kinetic energy at point B and the kinetic energy at point A is equal to the work done by the gravity between these two points

$Ec_B - Ec_A = m \cdot g \cdot (1 - a^2)$ , where $1-a^2$ is the y difference between points A and B

Sicne the bead starts at rest , the kinetic energy at point A is equal to 0 , while the kinetic energy at point B is equal to $\frac{m \cdot V^2}{2}$

After substituing , we get that $V = \sqrt{2 \cdot g \cdot (1 - a^2)}$

From here we can find Vx and Vy , but we don't know the angle t .

The direction of $\overrightarrow{V}$ is parallel to the line which is tangent to the graphic at point B $\Rightarrow$ $\tan t = y'(a) = 2 \cdot a$

Knowing that $\tan t = \frac{V_y}{V_x}$ and that $V_x^2 + V_y^2 = V^2$ we can find these two velocities in terms of a and V

Note that $V_x$ is constant throuought the air movement , while $V_y$ is not

Let $t_1$ be the time the bead takes to reach height h from the launching point , which is $t_1 = \frac{V_y}{g}$

Let $t_2$ be the time the bead takes to fall from height $a^2 + h$ , which is $t_2 = \sqrt{2 \cdot \frac{h+a^2}{g} }$ . where $h = \frac{V_y^2}{2 \cdot g}$

We know that the toatal distance from the origin $D =a + V_x \cdot (t_1 + t_2)$

After substituing and reducing everything possible we fing out that $D = a + \frac{2 \cdot a}{1 + 4 \cdot a^2} \cdot (2 - 2 \cdot a^2 + \sqrt{5 - 5 \cdot a^2 })$

The problem asks us to find a so that D is maximized . We take D as a function of a and find the roots of it's first derivative , which tells us that the maximum is attained when $\boxed{a = \frac{\sqrt{5}}{5}}$