A small ball is launched from a point $1.5\,\text{m}$ above level ground. When it reaches a height of $9\,\text{m}$ above the ground, it is moving at $45^\circ$ above the horizontal with speed $4\,\text{m s}^{-1}$.
(main)[4]
Determine the ball’s angle of projection.
Worked solution & mark scheme
This 4-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Horizontal component of motion: $V\cos\theta = 4\cos45$” …