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 begin by determining the horizontal and vertical components of the initial velocity.

1. Finding Horizontal Component of Velocity (U): The horizontal component of velocity can be found using the formula:

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

2. Finding the Vertical Component of Velocity (V): We need to calculate the vertical component of velocity when the ball is at a height of 3 m. Using the equations of motion:

   • Initial vertical velocity (v0) can be calculated from the total height:
     • Using the formula:
     $s = v_0t + \frac{1}{2}gt^2$
   • For the vertical motion, since g is downwards, we can express the vertical component at 2.5 seconds:
     • Rearranging gives us:
     $-10 = v_0(2.5) - \frac{1}{2}(9.81)(2.5^2)$
   • Solve for v0:
     • If we compute the above:
     • This gives:
     $v_0 = 3.067 \text{ m/s}$

3. Calculating Vertical Velocity (v) at height of 3 m:

   • Using the kinematic equation:

   $v^2 = v_0^2 + 2g h$

   • Substituting the values:

   $v^2 = (3.067)^2 + 2(-9.81)(3)$

   • Computing gives us:

   $v = 8.2625 \text{ m/s}$

4. Finding Total Speed:

   • Finally, the total speed at this height can be found from:

   $ext{Speed} = \sqrt{U^2 + V^2}$

   • Therefore,

   $ext{Speed} = \sqrt{16^2 + (8.2625)^2} = 18.61 \text{ m/s}$