<aside> 🚧 This document was not part of a session. It’s the explanation of the solution to a practice problem a learner asked on https://community.schoolhouse.world.
</aside>
Two frogs are sitting on separate lily pads in a pong. The first frog jumps from its lily pad with a maximum height of 60cm and lands in the water 20cm from the lily pad. The second frog is on a lily pad 10cm in front of the first lily pad. The second frog jumps at a maximum height of 40cm and lands in the water 40cm in front of its lily pad.
The first thing we need to do is to understand what the text is telling us and define a system of reference to solve the problem. The two frogs stand on two different starting lily pads, so we can put the centre of our system of reference in one of these two lily pads. I decided to define the first lily pad as point $(0, 0)$ and the second one as point $(10, 0)$ because they were easier to work with in my opinion compared to $(-10, 0)$ and $(0, 0)$.
A is the first lily pad and B is the second one
Then, we can draw the turning points of their trajectories. The two frogs jump up until a certain height and then come back to the ground at a certain distance from their starting point, so their top point will be $(\frac{distance\space covered}{2};max \space height)$, because the curve they draw is a parabola.
At is the top point of the first frog, Af its final point. Bt is the top point of the second frog, Bf its final point
Now, we need to find the equations of the trajectories that connect all three points through which each frog passes. Remember two things:
$$ A\begin{cases} 10=-\frac{b}{2a} \\ c=0 \\ 60=100a+10b+c \end{cases} $$
Let’s try to understand what’s going on here: 10 is the x-coordinate of the vertex of the parabola, so we’re setting it equal to the formula of the x-coordinate of the vertex. The term c is 0 because the parabola has a y-coordinate equal to 0 at x=0. The last equation is the equation of the vertex, where we set y=60 and x=10. Let’s solve this system of equations!
$$ A\begin{cases} 10=-\frac{b}{2a} \\ c=0 \\ 60=100a+10b+c \end{cases}
\begin{cases} b=-20a \\ c=0 \\ 60=-100a \end{cases}
\begin{cases} b=12 \\ c=0 \\ a=-\frac{3}{5} \end{cases} $$
This means that the equation of the first parabola is the following ⬇️
$$ \gamma_1:y=-\frac{3}{5}x^2+12x $$
We will take a similar approach to finding the second parabola, with the small difference that this time we don’t know the c term in advance so we will just use two point equations.