In this question use g = 9.81 m s^{-2} - AQA - A-Level Maths Mechanics - Question 17 - 2017 - Paper 2

To find the speed of the ball at a height of 3 metres, we first determine the horizontal and vertical components of the ball's motion.

1. Horizontal Velocity (U): The horizontal distance is given as 40 metres after 2.5 seconds:

   $U = \frac{\text{horizontal distance}}{\text{time}} = \frac{40}{2.5} = 16 \text{ m s}^{-1}$

2. Vertical Velocity (V): The vertical distance (y) can be expressed in terms of the initial vertical velocity (u) and time (t) using:

   $y = ut + \frac{1}{2}gt^2$

   Considering the downward direction as positive for gravity:

   $-10 = u(2.5) - \frac{1}{2}(9.81)(2.5)^2$

   Solving for u gives:

   $u = \frac{-10 + \frac{1}{2}(9.81)(2.5)^2}{2.5} = 8.2625 \text{ m s}^{-1}$

3. Calculating Speed at Height 3m: At a height of 3 m:

   By using:

   $V^2 = u^2 + 2g(y - y_0)$

   where ( y_0 = 0, g = -9.81 ), we have:

   $V^2 = (8.2625)^2 + 2(-9.81)(3)$

   Solving this will provide the speed of the ball when it is at 3 m above its initial position.