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WAEC 2016 · Paper 2 · Q15

  1. (a)

    A particle is projected vertically upwards from the ground with speed 30 m s−130\text{ m s}^{-1}. Calculate the: (i) maximum height reached by the particle; (ii) time taken by the particle to return to the ground; (iii) time(s) taken for the particle to attain a height of 40 m40\text{ m} above the ground. [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

  1. At the top v=0v = 0: 0=900−20s0 = 900 - 20s, so s=45 ms = 45\text{ m}.

(ii)

  1. Up: 0=30−10t0 = 30 - 10t, so t=3 st = 3\text{ s}.
  2. Coming down takes as long, so it lands after 6 s6\text{ s}.

(iii)

  1. 40=30t−5t240 = 30t - 5t^2, so t2−6t+8=0t^2 - 6t + 8 = 0.
  2. (t−2)(t−4)=0(t - 2)(t - 4) = 0: at t=2 st = 2\text{ s} and t=4 st = 4\text{ s}.

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