In this question use $g = 9.81 \, \text{m s}^{-2}$ - AQA - A-Level Maths Pure - Question 17 - 2017 - Paper 2

To find the speed of the ball at a height of 3 metres, we need to determine the components of the ball's initial velocity.

1. Finding Horizontal Component:
   The ball lands at the point $(40i - 10j)$ after 2.5 seconds. The horizontal distance traveled is 40m.

   The horizontal component of the initial velocity, $U_x$, can be computed as:

   $U_x = \frac{\text{Horizontal Distance}}{\text{Time}} = \frac{40}{2.5} = 16 \, \text{m s}^{-1}$

2. Finding Vertical Component:
   The vertical distance after 2.5 seconds is -10m. Therefore, using the equation of motion:

   $s = ut + \frac{1}{2}at^2$

   where:

   • $s = -10$ m
   • $u = U_y$ (vertical component of initial velocity)
   • $a = -g = -9.81 \, \text{m s}^{-2}$
   • $t = 2.5$ s

   Rearranging gives:

   $U_y = \frac{s - \frac{1}{2}at^2}{t} = \frac{-10 - \frac{1}{2}(-9.81)(2.5^2)}{2.5}$

   • After calculating, this results in:

   $U_y \approx 8.2625 \, \text{m s}^{-1}$

3. Finding speed at 3m height:
   To find the speed of the ball at a height of 3m, we use the vertical speed at this height. We use the formula for vertical motion:

   $v_y^2 = U_y^2 + 2as$ where $s = 3$ m (height above initial).

   Thus:

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

   Solving this gives:

   $v_y \approx 3.067 \, \text{m s}^{-1}$

4. Calculating total speed:
   Finally, the total speed $V$ at this height can be calculated:

   $V = \sqrt{U_x^2 + v_y^2} = \sqrt{16^2 + (3.067)^2} \approx 16.3 \, \text{m s}^{-1}$