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WAEC 2018 · Paper 2 · Q8

A body is thrown vertically upwards. Its height hh metres at time tt seconds is given by h=(12t−25t2) mh = \left(12t - \frac25t^2\right)\text{ m}. Find:

  1. (a)

    the time at which it is momentarily at rest;

  2. (b)

    the maximum height reached by the body.

Worked solution (try it first)

(a)

  1. The body is momentarily at rest when its velocity is zero: dhdt=12−45t=0\dfrac{dh}{dt} = 12 - \dfrac45t = 0.
  2. So 45t=12\dfrac45t = 12 and t=15t = 15 s.

(b)

  1. Put t=15t = 15 into the height: h=12(15)−25(15)2h = 12(15) - \frac25(15)^2
    =180−90= 180 - 90
    =90= 90 m.

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