A ball whooshes down a slide and hits another ball which flies off the slide horizontally as a projectile. How far does it go?
At what positions and speeds can the bomb be dropped to destroy the dam?
How fast would you have to throw a ball upwards so that it would never land?
In the problem Dam Busters 1 we looked at targetting a dam with a direct hit. In that case the plane could only just get close enough to the dam to destroy its target. The dam is destroyed if the bomb strikes above the rocky outcrop of height $B$. A scientist has suggested developing a bomb which will bounce on water when travelling transverse to the water surface at high speed. It has been suggested that this bouncing bomb will enable the target to be struck more centrally whilst releasing it from a greater distance. Several scientists have tried to calculate the height h of the bouncing bomb above sea level at the point at which it will be level with the wall after the first bounce. The scientists agree on the overall general form of the equation but cannot agree on some of the signs and which way around some of the factors should go. Can you decide whether each of the signs in the equation should be 'plus one' or 'minus one' (they can be different from each other)? $$ h = -\frac{1}{2}g^{\pm 1}\left[\left(\frac{V}{D}\right)^{\pm 1}-\sqrt{X}\right]^2 \pm eg\sqrt{X}\left[\left(\frac{V}{D}\right)^{\pm 1}-\sqrt{X}\right] \quad\quad X = \left(\frac{2H}{g}\right)^{\pm 1} $$ In the equation, e is the coefficient of restitution for the bomb striking the water. $e$ is dimensionless and positive. You do not need to derive the equation, only to decide which configurations would give rise to a sensible equation from a physical point of view.