A small ball is projected with speed U m/s from a point O at the top of a vertical cliff - Edexcel - A-Level Maths Mechanics - Question 5 - 2020 - Paper 1

To find U, we will use both horizontal and vertical motion equations.

1. Horizontal Motion: The horizontal distance traveled is given by:

   $x = U \cos(45^\circ) \cdot t$

   Setting this equal to 100 m, we have:

   $U \cos(45^\circ) \cdot t = 100$

   Thus, since ( \cos(45^\circ) = \frac{1}{\sqrt{2}} ):

   $U \cdot \frac{1}{\sqrt{2}} \cdot t = 100 \quad (1)$

2. Vertical Motion: The vertical motion can be described with:

   $h = U \sin(45^\circ) \cdot t - \frac{1}{2} g t^2$

   Here, h = -25 m (as it falls), with g approximated as 9.81 m/s², the equation becomes:

   $-25 = U \cdot \frac{1}{\sqrt{2}} \cdot t - \frac{1}{2} \cdot 9.81 \cdot t^2 \quad (2)$

3. Combining Equations: We can solve (1) to express t in terms of U:

   $t = \frac{100\sqrt{2}}{U} \quad (3)$

4. Substituting t: Substituting (3) into (2):

   $-25 = U \cdot \frac{1}{\sqrt{2}} \cdot \frac{100\sqrt{2}}{U} - \frac{1}{2} \cdot 9.81 \cdot \left( \frac{100\sqrt{2}}{U} \right)^2$

   After simplification, we get:

   $-25 = 100 - \frac{490.5 \cdot 200}{U^2}$

   Rearranging gives:

   $\frac{490.5 \cdot 200}{U^2} = 125$

   Finally solving yields:

   $U^2 = 4905 \Rightarrow U = 28 \text{ m/s}$