# Motion of tennis ball bouncing

I'm currently stuck on the last part of a question. Here is the diagram for the question: The last part of the question asks to calculate the direction of the ball at B. Assumptions are there is no air resistance and the ball bouncing does not affect the horizontal velocity of the ball. You are also given that the ball takes 0.6 seconds to travel the 24m and the height of B is 0.75m.

I know to solve the problem I need the horizontal and vertical components of the velocity at B and I can use trig to find the angle. The horizontal velocity I have correctly worked out to be 40m/s in a previous part of the question.

For the vertical velocity again in a previous part of the question I calculated the vertical velocity at A (correctly) to be 4.02m/s. I decided to calculate the impact velocity of the tennis ball as a starting point: \begin{align} v^2 =& u^2+2as \\ v =& \sqrt(4.02^2+2*9.8*2.8) \\ =& 8.43m/s \, . \end{align}

My assumption is that this is the initial vertical velocity as the tennis ball rises up off the ground. Working on that assumption: \begin{align} v =& u + at \\ =& 8.43-9.8*0.15 \\ =& 6.96m/s \end{align}

using $\tan\theta = 6.96/40$ gives $\theta = 9.87^\circ$.

The answer is supposed to be 6.09, I am fairly certain the 8.43 is correct, so where have I made a mistake?

• using $s=ut+1/2*at^2$ followed by $v=u+at$ and some trig did the trick. The reason I wasn't getting the correct answer when I had tried that method is because I forgot to include the $1/2$ in the first equation. So I wrongly wrote off that method which lead to my question above. – Darth Vader Jan 8 '17 at 16:47