(a)[2]
Show, by working, that $u = 22$.
(b)[4]
For $3.6\,\text{s}$, the particle is at a height greater than $h\,\text{m}$ above the ground. Find $h$.
Mathematics 9709 · AS & A Level · Kinematics of motion in a straight line
Kinematics of motion in a straight line — practice question
A particle is launched vertically upwards at speed $u\,\text{m s}^{-1}$ from a point on level ground. After $2$ seconds, its height above the ground is $24\,\text{m}$.
Worked solution & mark scheme
This 6-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Use of $s = ut + \frac{1}{2}at^2$, $24 = 2u - \frac{1}{2} g \times 2^2$” …
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