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

To find the speed when the ball is at a height of 3 metres, we first need to determine the components of its initial velocity.

1. Horizontal Motion: Using the horizontal distance covered and the time, we have:

   $U = \frac{40}{2.5} = 16 \, \text{m/s}$

2. Vertical Motion: The vertical displacement equation is given by:

   $s = ut + \frac{1}{2} a t^2$ Where:

   • s = vertical displacement
   • u = initial vertical velocity
   • a = acceleration due to gravity (downwards, hence -9.81 m/s²)
   • t = time (2.5 seconds)

   Setting $s = -10$ and solving for $u$:

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

   Rearranging gives: $u \approx 5.09 \, \text{m/s}$

3. Total Speed Calculation: Now, using the Pythagorean theorem to find the overall speed:

   $V = \sqrt{U^2 + V^2} = \sqrt{16^2 + 5.09^2} \approx 16.69 \, \text{m/s}$

   This is the speed when the ball is at a height of 3 metres.