(i)[4]
Find the value of $\theta$.
(ii)[4]
Show that after $4\,\text{s}$ the particle is $33.75\,\text{m}$ below the point of projection and determine the direction of motion at that moment.
Mathematics 9709 · AS & A Level · Probability
Probability — practice question
A particle is launched at speed $15\text{ m s}^{-1}$ at an angle of $\theta^\circ$ above the horizontal. Four seconds after launch, the particle's speed is $30\text{ m s}^{-1}$.
Worked solution & mark scheme
This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Horizontal component is $15\cos\theta=v_H$ and vertical component is $15\sin\theta-4g=v_V$” …
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