Pythagoras - time/speed/distance

Question

In a game of tennis, a ball leaves a racket in a horizontal direction from a height of 0.39 m above the ground. The speed of the ball is 35 ms-1. The ball takes 0.28 seconds to reach the ground again. What horizontal distance has the ball travelled in this time?

So I took this as:

Horizontal = √ (35x0.28)² - (0.39)²

= 9.79m (to 2d.p.) - full answer is 9.792236721

However, my answer key says 9.90m - why?

Breakdown:

I create a right-angled triangle:

A = racket height

B = floor

C = where the ball hits ground

To get the ball trajectory (AC) I did 35m/s x 0.28s = 9.8m (this being the hypotenuse)

Floor to racket (AB) = 0.39

So I am looking for BC right?

So with pythag surely it's:

(AC)² - (AB)² = (BC)²

9.8² - 0.39² = (BC²)

96.04 - 0.1521 = 95.8879 (BC²)

Sqrt 95.8879 (BC²) = 9.79 (BC)

However, my answer key says 9.90m - why?

How do I get to 9.90m so I get the mark on the exam for a similar question?

Answer 1

I think your answer is correct, i.e. I do not see an error in your solution.