In this question use g = 9.81 m s² - AQA - A-Level Maths Pure - Question 17 - 2017 - Paper 2

To find the speed of the ball when it is at a height of 3 metres, we first need to calculate the horizontal and vertical components of the initial velocity.

1. Horizontal Component of Initial Velocity: The horizontal position after 2.5 seconds is given as 40 m. Thus, the horizontal component of the initial velocity ( U_x) is:

   $U_x = \frac{40 ext{ m}}{2.5 ext{ s}} = 16 ext{ m s}^{-1}$

2. Vertical Component of Initial Velocity: Using the vertical displacement equation:

   $S_y = U_y t + \frac{1}{2} a t^2$ We know that:

   • Displacement (S_y) is -10 m (as it falls below the starting point)
   • Time (t) is 2.5 s
   • Acceleration (a) is -9.81 m s²

   Plugging in the known values:

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

   Solving for U_y gives:

   $U_y = \frac{-10 + 30.625}{2.5} = 8.2625 ext{ m s}^{-1}$

3. Speed at 3 Metres Height: Next, we calculate the velocity at the height using the conservation of mechanical energy or kinematic equations. At 3 m height, using:

   $v_y^2 = U_y^2 + 2aS \rightarrow v_y^2 = (8.2625)^2 + 2(-9.81)(3)$ Applying: $v_y^2 = 68.307 + (-58.86) = 9.447$ $v_y = \sqrt{9.447} = 3.07 ext{ m s}^{-1}$

4. Total Speed: Now, the total speed ( V) is given by:

   $V = \sqrt{U_x^2 + v_y^2} = \sqrt{(16)^2 + (3.07)^2} = \sqrt{256 + 9.42} = \sqrt{265.42} \approx 16.3 ext{ m s}^{-1}$