A simplified diagram of the situation reveals that the problem is about similar triangles.

(i) Since the two triangles are similar, the fraction $0.9/y = 5/15$

Rearranging the equality gives $y$ = $2.7$ metres.

So the minimum height from which the ball can be hit from is $2.7$ metres.

(ii) Using Pythagoras's Theorem $(a^2 + b^2 = c^2)$ we can work out how far the ball travels before it hits the ground.

$2.7^2 + 15^2 = x^2$

$x = \sqrt{232.29}$

$x = 15.24$ metres

So the distance travelled by the ball before it hits the ground is $15.24$ metres.

(iii) For the ball to travel the shortest distance (and hence the shortest time) it should be hit so that it bounces on the base line, however this serve would be a fault. To give the receiver the least time to return the serve, the ball would have to bounce on the service line. In this situation the height to which the ball rises after it has bounced will be a minimum.

How far would the ball travel before it was hit back?