Maths prize 17 November 2017: ball and bucket

Maths prize 17 November 2017: ball and bucket

Ashley drops a bucket from a height h above the ground. At the same moment Brenda, at ground level and at a horizontal distance s from the falling bucket, throws a ball with speed v at exactly the right angle θ to hit the bucket as it falls. Brenda’s height is negligible compared to h.

Find θ, the time t at which the ball hits the bucket, and the minimum v to be able to hit the bucket before it reaches the ground.

The Year 13 Further Maths class, collectively, with Tegan Hill doing much of the work, worked out how to find θ and t, and Mohaned al-Bassam at least started an attempt to find v. Jeffrey Sylvester made a good effort. But, disappointingly, no-one wrote up an attempted solution.

Take a simple special case: if g is negligibly small. Then:

\theta = \arctan \frac{h}{s}

The ball hits the bucket after it has dropped only a negligible distance.

Imagine g increases. The bucket falls, and the ball falls off vertically from the trajectory it would have with negligible g, both with the same acceleration g, and over the same time as each other. Therefore, as g increases from negligible amounts, the required angle θ remains the same, because the falling-down and the falling-off increase exactly in line with each other. Therefore, whatever g:

\theta = \arctan \frac{h}{s}

The time t= time taken by ball at horizontal speed v cos θ to travel distance s, so, given v, h, s , does not depend on g. So it is the same as it would be if g were negligible:

t = \frac{\sqrt{h^2+s^2}}{v}

For the ball to hit the bucket before the ground,

\frac{1}{2}gt^2 \leq h , so:

\frac{g(h^2+s^2)}{2v^2} \leq h

v \geq \sqrt{\frac{g(h^2+s^2)}{2h}}

For a much less neat way of doing this, see: