Spaceballs (Part 2)

Refer to Spaceballs (Part 1) to derive the graph that both balls follow on the xy-plane here .

$y=\frac{-x^{2}}{40}$

To find how much time the ball spends in the air, we first need to find when it will land again. We do this by defining the plank's position as $y=-x$ on our xy-plane. Now, we need to see when the ball's path will hit the plank.

$-x=\frac{-x^{2}}{40}$ $x=40m$ $y=-40m$

Now that we know the ball's x- and y-coordinates when it reaches its second bounce, we can solve for how long it took the ball to travel those distances.

$40=14\times t_{1}$ $t_{1}=\frac{20}{7}s$ $-40=\frac{-9.8}{2}t_{1}^2$ $t_{1}=\frac{20}{7}s$

Since we got that it took $\frac{20}{7}s$ for the ball to reach the plank using both the x- and the y-coordinates, we can be confident that the first part of our answer is correct. The next part is to find out how much time it took for the ball to drop $10m$ after being released from rest.

$-10=\frac{-9.8}{2}t_{2}^2$ $t_{2}=\frac{10}{7}s$

Finally, we add $t_{1}$ and $t_{2}$ to get $\boxed{\frac{30}{7}s}$ .